PRINCIPLES 

OF 

ELECTRICAL  ENGINEERING 


BY 

WILLIAM  H.  TIMBIE 

AND 

VANNEVAR  BUSH 

Associate  Professors  of  Electrical  Engineering 
Massachusetts  Institute  of  Technology 


TOTAL   ISSUE,    FOUR   THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:   CHAPMAN  &  HALL,   LIMITED 
1922 


COPYRIGHT,  1922 

BY 
WILLIAM  H.   TIMBIE 

AND 

VANNEVAR  BUSH 


TECHNICAL    COMPOSITION    CO. 
CAMBRIDGE,    MASS.,    U.  S.  A. 


PREFACE 

This  text  is  the  outgrowth  of  experience  in  teaching  the 
principles  of  electrical  engineering  to  students  of  electrical 
engineering  at  the  Massachusetts  Institute  of  Technology. 
It  aims  to  provide  a  substantial  first  course  in  the  subject 
by  presenting  rigorously,  and  at  the  same  time  in  under- 
standable form,  the  really  basic  principles  upon  which 
modern  electrical  engineering  rests.  In  furtherance  of  this 
purpose  many  problems  and  examples  from  current  engineer- 
ing practice  are  introduced.  The  book  is  not,  however, 
to  be  mistaken  for  a  complete  condensed  treatise  on  the 
entire  subject.  It  is  strictly  a  first  course  on  the  principles, 
and  its  study  should  be  followed  by  detailed  courses  in 
direct-current  and  alternating-current  machinery.  Where- 
ever  applications  of  the  principles  are  introduced,  they  are 
for  the  purpose  of  illustrating  these  principles  and  render- 
ing them  real  and  alive  to  the  student. 

The  book  has  the  following  special  features,  which  we 
believe  to  be  desirable: 

1.  The  subject  of  the  magnetic  circuit  has  been  stressed. 
It  has  been  the  common  experience  of  teachers  of  electrical 
engineering  that  students  beginning  the  subject  find  this  a 
stumbling   block.     Much   more   space   than  is   usual   has, 
therefore,  been  devoted  to  this  matter. 

2.  As  a  basis  for  explanation,  the  modern  electron  theory 
has  been  freely  used.     It  has  been  found  that  this  affords 
the  most  rational  means  of  tying  together  the  otherwise 
widely  divergent  principles  with  which  the  electrical  engi- 
neer deals. 

3.  The    subjects    of    thermionic    emission,    conduction 
through   gases,    electrolytic    conduction   and   certain   high- 
frequency  phenomena  have  been  included.    A  knowledge 

iii 


IV  PREFACE 

of  these  matters  is  becoming  more  and  more  essential  to 
the  electrical  engineer  in  any  field. 

4.  The  subject  of  the  behavior  of  dielectrics  has  been 
approached   from   a   standpoint   which   departs   from   the 
historical  method  of  attack  in  order  to  gain  clearness  and 
unity  of  treatment. 

5.  About  five  hundred  live  problems  are  included  for 
illustration,  for  practice  in  applying  the  principles  and  for 
the  purpose  of  bringing  before  the  student  useful  and  in- 
teresting engineering  data.     Some  of  these  are  purposely 
made  of  such  calibre  as  to  merit  the  attention  of  the  most 
able  students. 

The  book  is  written  primarily  for  students  of  college 
grade  and  presumes  a  knowledge  of  calculus  and  physics. 
The  terminology  and  symbols  employed  are  those  recom- 
mended by  the  American  Institute  of  Electrical  Engineers. 

Grateful  acknowledgment  is  extended  to  Professor  F.  S. 
Dellenbaugh,  Jr.,  for  oscillographs  of  transients,  and  to 
Mr.  E.  L.  Bowles,  Mr.  L.  F.  Woodruff  and  Mr.  E.  L.  Rose 
for  diagrams,  proof-reading  and  the  checking  of  problems. 

W.  H.  T. 
V.B. 

CAMBRIDGE,  MASSACHUSETTS, 
February,  1922. 


TABLE  OF  CONTENTS 

CHAPTER  I 

THE  ELECTRICAL  ENGINEER  PAGE 

Electricity  not  a  Natural  Source  of  Power  —  Why  We  Have 
Central  Power  Plants  —  Why  Central  Power  Plants  are 
Electrical  —  Locations  of  Power  Plants  —  Convenience  of 
Electrical  Power  —  The  Electrical  Engineer 1 

CHAPTER  II 

ELECTRIC  UNITS  AND  ELECTRIC  CIRCUITS 

The  Electron  Theory  —  The  Ampere,  a  Unit  of  Current  —  The 
Volt,  a  Unit  of  Pressure  —  The  Ohm,  a  Unit  of  Resistance  — 
Ohm's  Law  —  Absolute  System  of  Units:  The  Abvolt, 
Abampere  and  Abohm  —  General  Application  of  Ohm's 
Law  —  Kirchhoff 's  Laws  —  Electromotive  Force  and  IR 
Drop  —  Special  Networks,  the  Delta  and  the  Star  —  Meas- 
urement of  Current,  Voltage  and  Resistance  —  The  Wheat- 
stone  Bridge ! 17 

CHAPTER  III 
ELECTRIC  POWER  AND  ENERGY 

The  Power  Equation  —  Voltage  not  a  Force  —  Use  of  the  Power 
Equation  —  Power  Consumed  by  Resistance  —  Measure- 
ment of  Electric  Power  —  Electric  Energy  —  Heat  Energy  of 
Electricity  —  Efficiency  of  Transmission,  Regulation  —  The 
Three- Wire  System  of  Transmission 52 

CHAPTER  IV 

THE  COMPUTATION  OF  RESISTANCE 

Resistivity  —  Resistance  per  Mil-foot  —  Resistivity  of  Metals 
used  as  Electrical  Conductors  —  Conductivity  of  Materials 
—  Temperature  Coefficient  of  Resistance  —  Temperature 
Change  Measured  by  Change  in  Resistance  —  Temperature 
Coefficient  of  Alloys,  etc.  —  Copper- Wire  Tables  —  Stranded 


VI  TABLE  OF  CONTENTS 

PAGE 

Wire  —  Aluminum  —  Copper-Clad  Steel  Wire  —  Safe  Car- 
rying Capacity  of  Wires  —  Determination  of  Right  Sizes  for 
Interior  Wiring  —  Insulating  Materials  —  Insulation  Re- 
sistance of  Cables 76 

CHAPTER  V 
ELECTROLYTIC  CONDUCTION 

Electrolytes  and  lonization  —  Electrolytes  and  Dissociation  — 
Electric  Potential  Series  —  Quantity  Relations  —  Primary 
Cells  —  Storage  Batteries  —  Electrolytic  Refining  of  Metals 

—  Electrolysis  —  Other   Electrochemical   Processes 107 

CHAPTER  VI 
THE  MAGNETIC  CIRCUIT 

Relation  between  Electricity  and  Magnetism  —  The  Magnetic 
Circuit  —  Measurement  of  Flux  —  Flux  Lines  —  Magneto- 
motive Force  —  Ohm's  Law  for  Magnetic  Circuits  —  Re- 
luctances in  Series  and  in  Parallel  —  Variation  of  Permea- 
bility —  Ampere-turns  to  Produce  a  Given  Flux  —  Flux 
Produced  by  a  Given  number  of  Ampere-turns  —  Air  Gaps 

—  Leakage  Flux 136 

CHAPTER  VII 
THE  MAGNETIC  FIELD 

The  Line-Integral  Law  —  Field  About  a  Long  Wire  —  Field  In- 
side a  Conductor  —  Flux  Density  at  the  Center  of  a  Single 
Turn  of  Circular  Form  Carrying  Current  —  Flux  Density 
at  a  Point  on  the  Axis  of  a  Circular  Coil  —  The  Air-Core 
Solenoid  —  Calibration  of  a  Ballistic  Galvanometer  —  The 
Toroid  —  Flux  about  a  Conductor  in  a  Slot  —  Introduction 
of  Iron  into  a  Magnetic  Field  —  Magnetic  Poles  and  Pole 
Strength 183 

CHAPTER  VIII 
INDUCED  VOLTAGES 

Change  of  Linkages:    Lena's  Law  —  Self-induction    Coefficient 

—  Transients  in  Inductive  Circuits  —  Time  Constant  —  In- 
ertia of  an  Electric  Circuit  —  Energy  of  a  Magnetic  Field  — 
Magnetic  Pull  —  Mutual  Induction 250 


TABLE  OF  CONTENTS  vii 

CHAPTER  IX 

MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL  PAGE 

Magnetic  Retentivity  and  Hysteresis  —  Energy  of  Magnetic 
Field  in  Iron  —  Hysteresis  Loops  —  Mean  Magnetization 
Curves:  Froelich's  Equation  —  Methods  of  Demagnet- 
izing Steel  —  The  Steinmetz  Equation  —  Effect  of  the 
Composition  of  Steel  —  Permanent  Magnets 298 

CHAPTER  X 
GENERATED  VOLTAGE 

Change  of  Linkages  —  Elementary  Alternators  —  The  Direct- 
Current  Generator  —  Conductor  in  a  Moving  Field  —  The 
Homopolar  Generator  —  Eddy  Currents 328 


CHAPTER  XI 
FORCE  ON  A  CONDUCTOR 
Force  on  a  Conductor  Carrying  a  Current  —  Meters  —  Motors 

—  The  Back  Electromotive  Force  —  Speed 367 

CHAPTER  XII 
CONDUCTION  THROUGH  GASES 

Non-Metallic  Conduction  of  Electricity  —  Thermionic  Con- 
duction—  Richardson's  Law  of  Thermionic  Conduction  — 
Thermionic  Amplifiers  and  Oscillators  —  X-Ray  Tubes  — 
The  Gaseous  Discharge  Tube  — The  Spark  —  Corona  — 
Arcs 391 

CHAPTER  XIII 

DIELECTRICS 
Dielectric  Strength  —  Condenser  Action  —  Dielectric  Constant 

—  Parallel-Plate  Condensers  —  Charge  on  a  Condenser  — 
Measurement    of     Capacitance  —  Charging    a     Condenser 
through   a   Resistance  —  Discharge  —  Energy    Relations  — 
Mechanical  Force  in  a  Condenser  —  Electrostatic  Fields  — 
Capacitance  of  a  Pair  of  Long  Aerial  Conductors  —  Ap- 
plication of  a  Sinusoidal  Voltage  to  a  Condenser  —  Con- 
denser Losses:  Dielectric  Hysteresis  —  Condensers  in  Par- 
allel and  Series  —  Distribution  of  Stress  in  Insulation  — 
Cylindrical  Condenser 431 


Vlll  TABLE  OF  CONTENTS 

APPENDIX  PAGE 

Table  I,  Resistivity  and  Temperature  Coefficient  —  Table  II, 
Resistance  of  International  Standard  Annealed  Copper  — 
Table  III,  Allowable  Carrying  Capacities  of  Wires  —  Table 
IV,  Atomic  Weights  and  Usual  Valences  of  Some  Elements 
—  Table  V,  Specific  Heat  of  Various  Materials  —  Table  VI, 
Dielectric  Constants . .  494 


Principles  of 
Electrical  Engineering 

CHAPTER   I 
THE   ELECTRICAL   ENGINEER 

A  nation's  standing  in  the  scale  of  modern  civilization 
can  fairly  be  measured  by  the  extent  to  which  she  is  utilizing 
her  natural  sources  of  power.  The  truth  of  this  is  evident 
in  the  case  of  manufacturing  nations,  but  nowadays  it  is 
also  true  of  even  agricultural  communities  since  they  de- 
pend for  much  of  their  prosperity  upon  the  artificial  fer- 
tilizer produced  electrically  by  plants  utilizing  natural  water 
power  that  formerly  went  to  waste. 

This  development  of  the  world's  sources  of  power  is  the 
work  of  the  engineer.  Whether  the  task  is  the  designing  of 
some  delicate  mechanism  to  operate  on  the  minute  quantity 
of  power  available  in  a  telephone  circuit,  or  the  harnessing 
of  the  floods  of  the  Mississippi  to  supply  whole  cities  with 
almost  unlimited  power,  it  is  the  task  of  the  engineer. 
Wherever  a  project  involves  the  generation,  transmission 
or  utilization  of  electric  power,  electrical  engineers  are  needed. 

1.  Electricity  not  a  Natural  Source  of  Power.  Elec- 
tricity as  it  occurs  in  nature,  however,  is  not  one  of  the 
natural  sources  of  power,  although  fully  one-half  the  world's 
power  supply  is  converted  into  the  electrical  form  before  it 
is  finally  utilized  because  that  is  often  the  cheapest  and  most 
convenient  form  in  which  to  transmit  and  utilize  power. 

1 


^'PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


chief  natural  sources  of  power  are  : 

(1)  Water  flowing  in  natural  or  artificial  waterways, 

(2)  Coal  which  is  mined  from  the  earth, 

(3)  Oil  and  gas  which  flow  from  wells. 

In  order  to  generate  any  large  amount  of  power  by  means 
of  water,  either  there  must  be  an  enormous  flow  of  water  at 
a  low  head  or  a  smaller  flow  at  a  very  high  head.  Thus  in 
the  plant  at  Swan  Falls,  Idaho,  which  is  shown  in  Fig.  1, 
the  head  averages  only  19  feet  and  the  necessary  flow  per 
horse  power  is  over  120,000  pounds  per  hour,  while  in  the 
San  Franciscito  plant  in  California,  shown  in  Fig.  2,  the 
flow  per  horse  power  is  less  than  2600  pounds  per  hour  but 
the  head  is  938  feet.  The  design  of  the  water  wheels  and 
electric  generators  for  operating  under  such  diverse  con- 
ditions must  show  a  wide  variance.  In  a  problem  of  this 
kind,  the  electrical  engineer's  knowledge  of  physics,  par- 
ticularly mechanics,  is  called  most  prominently  into  play. 
Where  coal  or  oil  is  used  to  produce  power,  a  much  smaller 
quantity  per  minute  is  required.  For  instance,  in  place  of 
either  of  the  two  projects  mentioned  above,  we  could  pro- 
duce a  horse  power  by  burning  about  1.5  pounds  of  coal  or 
0.9  pound  of  oil  per  hour.  Thus  by  burning  a  small 
quantity  of  oil  or  coal,  an  immense  amount  of  energy  can 
be  released.  This  is  one  of  the  reasons  why  the  larger 
percentage  of  the  power  used  in  the  world  comes  from  coal 
and  oil,  in  spite  of  the  fact  that  the  sources  of  these  materials 
are  limited  in  extent. 

2.  Why  We  have  Central  Power  Plants.  Great  as  the 
amount  of  power  is  that  can  be  obtained  from  small  amounts 
of  oil  and  coal,  it  is,  however,  rarely  ever  convenient  or 
economical  to  set  up  an  engine  at  just  the  place  where  the 
power  is  to  be  used.  When  we  wish  to  light  a  room  by  in- 
candescent electric  lamps  which  require  about  ^  horse 
power  each,  we  do  not  set  up  a  small  gasoline  motor  and 
generator  or  a  steam  engine  and  generator  near  each  lamp. 
Neither  do  we  set  up  in  a  shop  a  small  engine]  near  each 


THE  ELECTRICAL  ENGINEER 


PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


$* 

h  <n 


fc>D   o 

.s  -* 

.2     03 


THE  ELECTRICAL  ENGINEER  5 

machine  to  be  driven.  In  fact,  in  a  large  manufacturing 
plant,  we  rarely  have  separate  engines  for  separate  build- 
ings. One  central  plant  supplies  several  buildings.  A 
further  development  of  this  same  scheme  is  to  supply  power 
to  a  whole  city  or  community  from  a  single  power  plant. 
Recently  power  plants  have  been  developed  which  supply 
many  cities  at  once.  The  modern  tendency  is  to  include 
larger  and  larger  areas  within  one  system,  some  engineers 
even  prophesying  that  all  the  power  stations  in  the  whole 
country  will  eventually  be  connected  into  one  vast  super- 
power system. 

3.  Why  Central  Power  Plants  are  Electrical.  When  it 
is  desired  to  transmit  power  over  great  distances,  as  in  the 
case  of  the  systems  mentioned  above,  practically  the  only 
way  in  which  it  can  be  done  effectively  is  by  means  of 
electricity.  In  fact,  when  power  is  to  be  transmitted  more 
than  a  few  feet,  electricity  is  usually  the  best  form.  Con- 
sider what  a  single  room  of  an  old-fashioned  shop  looked 
like  with  its  forest  of  belts  and  shafts  and  pulleys.  Then 
call  to  mind  the  picture  that  would  be  presented  by  a  whole 
building  or  a  group  of  buildings  connected  by  rapidly  moving 
belts  and  ropes.  Try  to  imagine  what  even  a  small  town 
would  look  like  if  we  tried  to  transmit  power  from  a  central 
plant  to  the  various  manufacturing  plants  by  means  of  belts 
and  ropes,  and  it  will  be  evide£t  why  central  power  stations 
are  electrical.* 

The  advantages  may  be  summed  up  as  follows: 
1.   Electrical  energy  is  transmitted  by  wires  which  are 
small  and  cheap,  which  do  not  move,  which  can  be  bent  into 

*  It  must  not  be  inferred  from  this  that  all  other  means  of  power 
transmission  could  or  should  be  superseded  by  electrical  devices. 
Where  distances  are  short,  belts  and  shafting  are  often  cheapest  and 
most  efficient.  Where  installations  are  temporary,  distances  not  too 
long  and  only  small  quantities  of  power  are  needed,  as  for  rock  drills, 
hoisting  engines  and  pumps,  steam  under  pressure  may  often  be  used  to 
advantage.  Compressed  air  is  also  often  preferable  under  some 
conditions. 


6          PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

almost  any  shape  to  pass  around  curves  and  obstacles  and 
which  are  comparatively  safe,  neat  and  of  small  upkeep 
expense  when  once  installed. 

2.  Electrical  energy  can  be  converted  easily  into  nearly 
any  other  form  at  the  spot  where  it  is  needed.     It  lights 
lamps,   drives  motors,  heats  stoves  and  furnaces,   refines 
metals  and  electroplates. 

3.  Very  little  energy  is  lost  in  the  wires  and  other  devices 
during  transmission  over  even  great  distances. 

4.  Electrical  devices  are  easily  stopped,  started  and  con- 
trolled by  simple  compact  devices  which  are  rapid  and  ac- 
curate in  their  operation  and  durable  in  their  construction. 

4.  Locations  of  Power  Plants.  For  the  purpose  of  econ- 
omical distribution,  a  power  plant  should  be  located  as 
near  as  possible  to  the  center  of  the  district  it  supplies.  In 
the  case  of  plants  operated  by  water,  such  a  location  is 
generally  impossible.  Such  large  quantities  of  water  under 
great  pressure  are  needed  for  most  plants  that  it  is  more 
practicable  to  locate~the  plant  near  the  water  way  and 
transmit  the  electrical  power  to  the  district  to  be  supplied 
than  to  bring  so  much  water  to  a  centrally  located  plant. 

The  problem  of  central  locations  of  plants  operated  by 
gas,  oil  or  coal  is  much  simpler.  The  quantity  of  energy 
contained  in  a  small  weight  of  gas,  oil  or  coal,  the  ease 
with  which  gas  and  oil  can  be  piped  over  great  distances 
and  the  numerous  facilities  for  transporting  coal  make  it 
practicable  to  locate  the  power  plant  at  the  center  of  the 
district  to  be  served. 

In  the  future,  however,  as  the  efficiency  of  transmitting 
electrical  energy  becomes  greater  and  the  cost  of  transport- 
ing gas,  oil  and  coal  becomes  higher,  it  may  be  more  econom- 
ical to  locate  even  the  plants  using  these  fuels  in  the  oil  and 
coal  fields  and  transmit  power  by  electricity  to  still  greater 
distances  than  at  present.* 

*  See  "Electrical  Transmission  versus  Coal  Transportation",  by 
Harold  W.  Smith,  Electrical  Journal,  September,  1921. 


THE  ELECTRICAL  ENGINEER  7 

5.   Superpower   System.     The   great  saving   in   cost   of 
energy  and  in  conservation  of  coal  and  oil  made  possible 


•New  York^. 

PROPOSED 
SUPERPOWER  TRANSMISSION  SYSTEM 

Interconnecting  Network . 

Reinforcements  to  1.  N.  Due  to  St  Lawrence . 


_____  Proposed  Trans.  Lines  From  St.  La» 
_„__._  Trans.  Lines  From  Remote  Sources. 
From  Niagara  Fall*. 

•  Hydro-Electric  Plants. 

A  Steam-Electric  Plants. 

A  Load  Centers  With  Steam-Elec.  Plants. 

e  Load  Centers. 

m  Switching  Stations  not  at  Load  Cent. 


FIG.  3.  It  is  proposed  to  develop  all  the  water  power  and  inter- 
connect all  the  power  plants  within  this  section,  which  already 
carries  on  forty  per  cent  of  the  manufacturing  of  the  United  States. 

by  developing  a  superpower  system  are  strikingly  illus- 
trated in  the  Report  of  the  Superpower  Survey.* 

*  Professional  Paper  No.  123,  Department  of  Interior,  U.  S.  Geological 
Survey,  entitled  "A  Superpower  System  for  the  region  between  Boston 
and  Washington,"  by  W.  S.  Murray  and  others. 


8          PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

It  will  be  noted  from  Fig.  3  that  the  Superpower  Zone 
extends  from  Washington,  D.  C.,  to  Portland,  Me.,  and 
from  the  Atlantic  seaboard  to  Harrisburg,  Pa.,  and  Utica, 
N.  Y.  Within  this  zone  reside  twenty-five  millions  of  the 
nation's  population  and  here  forty  percent  of  the  manufac- 
turing of  the  nation  is  carried  on.  There  are  already  in  this 
area  315  electric  public  utilities  operating  about  1200  miles 
of  line  at  33,000  volts  or  higher,  to  which  the  superpower 
system  will  add  about  4700  at  110,000  volts  and  4000  at 
220,000  volts.  While  most  of  the  power  is  to  be  developed 
from  the  coal,  oil  and  water  power  within  this  territory,  it  is 
also  proposed  to  draw  huge  quantities  from  an  enlarged 
power  development  at  Niagara  Falls  and  a  new  plant  to  be 
established  on  the  St.  Lawrence  River.  All  of  the  avail- 
able water  power  within  the  area  will  be  put  to  use,  the 
Potomac,  Susquehanna  and  Delaware  Rivers  furnishing 
power  to  the  central  and  southern  portions,  the  Hudson 
and  Connecticut  Rivers  to  the  northern  portion.  How- 
ever, these  water-power  developments  even  when  sup- 
plemented by  those  of  Niagara  and  the  St.  Lawrence  will 
take  care  of  but  little  more  than  twenty  percent  of  the 
requirements  of  the  territory  under  consideration.  For 
about  eighty  percent  of  the  power  in  this  vast  system,  coal 
and  oil  must  be  depended  upon,  and  it  is  in  the  economical 
location  and  operation  of  these  steam-electric  stations  that 
the  great  saving  is  to  be  accomplished.  Note  that  these  are 
to  be  located  in  the  coal  fields  at  Sunbury,  Nescopee  and 
Pittston,  Pa.,  and  at  cities  on  the  tidewater  to  which  trans- 
portation is  easy  and  cheap. 

Most  of  this  development  bids  fair  to  take  place  within 
ten  years.  One  of  the  problems  involves  the  designing  of  ap- 
paratus with  an  insulation  capable  of  withstanding  about  a 
half -million  volts  pressure.  A  knowledge  of  the  actions  of  dif- 
ferent materials  when  subjected  to  this  pressure  is  necessary, 
and  for  this  reason  the  subject  of  electrostatics  must  be 
mastered  by  an  engineer  entering  this  part  of  the  electrical 


THE  ELECTRICAL  ENGINEER  9 

field.  To  this  must  be  added  a  knowledge  of  the  laws  of 
economics,  for  the  relative  costs  of  different  materials, 
methods  and  systems  are  always  a  vital  factor  in  all  great 
engineering  projects. 

6.  Convenience  of  Electric  Power.     It  is  not  only  in  the 
economical  generation,  transmission  and  utilization  of  huge 
quantities  of  power  that  electricity  excels.     It  is  also  in  the 
number  of  conveniences  made  possible  by  its  use.     Chief 
among  these  is  the  telephone.     Each  instrument  requires 
only  a  minute  quantity  of  power  but  makes  possible  com- 
munication with  any  of  thirteen  million  stations  in  the 
United  States.     It  has  long  been  possible  to  carry  on  a  con- 
versation between  New  York  and  San  Francisco  over  metal- 
lic   conductors.     Recently    speech    was    transmitted    from 
Havana  to  Key  West  by  submarine  cable,  across  the  con- 
tinent to  San  Francisco  by  aerial  wire  and  underground  cable, 
and  over  a  thirty-mile  strip  of  ocean  to  Catalina  Island  by 
radio.     This   accomplishment   of   combining   wireless   tele- 
phony with  the  present  wire  systems,  with  their  twenty-five 
million  miles  of  conductors,  gives  promise  of  far-reaching 
development  in  the  near  future. 

The  cross-continent  transmission  was  made  possible  by 
vacuum-tube  repeaters  inserted  in  the  line.  The  wireless 
transmission  was  accomplished  through  vacuum  tubes  used 
as  rectifiers  and  amplifiers.  In  fact,  the  vacuum  tube  has 
recently  come  to  play  a  prominent  part  in  so  many  electrical 
developments  that  a  knowledge  of  vacuum  apparatus  and 
the  conduction  of  electricity  through  gases  is  almost  a 
prime  requisite  for  a  modern  electrical  engineer  in  any  field. 

7.  The    Electrical    Engineer.     The    two    examples   just 
given  of  progress  in  electrical  engineering,  the  superpower 
project  and  long-distance  communication,  were  chosen  be- 
cause of  the  great  difference  in  their  respective  fields.     One 
deals  with  vast  quantities  of  power  at  high  voltages,  capable 
of  being  applied  to  thousands  of  different  uses.     The  other 
employs  insignificant  amounts  of  power  at  the  lowest  volt- 


10        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

ages  and  its  use  is  restricted  to  the  carrying  on  of  conver- 
sation over  distances.  Yet  both  of  these  enterprises  re- 
quire the  services  of  exactly  the  same  types  of  electrical 
engineers  with  the  same  knowledge  of  fundamental  laws  of 
electricity  and  the  same  training  in  applying  them. 

Electrical  engineers  may  be  divided  into  three  more  or 
less  clearly  defined  classes  or  types,  depending  upon  their 
duties. 

First  type,  —  the  electrical  engineer  who  applies  the  laws 
of  science  to  the  development  of  electrical  equipment. 
To  this  class  belongs  the  research  engineer,  who  pushes 
outward  the  boundaries  of  our  knowledge  of  natural  laws 
for  the  purpose  of  employing  them  more  fully  for  mankind, 
the  designing  engineer, who  plans  the  machines  which  make 
use  of  these  lawsiand  the  production  engineer,  who  manu- 
factures them.  Each  must  have  a  thorough,  up-to-date 
knowledge  of  the  main  branches  of  physics  and  chemistry 
as  well  as  an  intimate  knowledge  of  the  latest  theories  and 
discoveries  in  electrical  science.  The  last  two  must  also 
be  familiar  with  the  best  shop  methods  especially  as  related 
to  quantity  production.  To  this  class  of  engineers  all  elec- 
trical projects  must  look  to  supply  the  necessary  electrical 
machinery  and  appliances. 

Second  type,  —  the  electrical  engineer  who  applies  the  elec- 
trical-engineering equipment  to  the  use  of  man.  He  plans, 
constructs  and  operates  power-transmission  systems,  tele- 
phone, telegraph  and  electric-railway  systems.  To  this 
class  belongs  the  engineer  who  has  the  responsibility  of 
selecting  the  proper  types  and  sizes  of  electrical  equipment. 
He  must  be  familiar  with  the  latest  designs  and  must  have 
all  the  fundamentals  of  the  science  so  well  in  hand  as  to  be 
able  to  judge  fairly  the  claims  of  various  manufacturers 
concerning  their  outputs.  Closely  connected  with  this  task 
is  the  writing  and  interpreting  of  specifications.  Such  an 
engineer  must  not  only  have  a  knowledge  of  the  standard 
tests  and  behavior  of  electrical  machinery  under  given  con- 


THE  ELECTRICAL  ENGINEER  11 

ditions  but  must  also  be  able  to  describe  these  phenomena  in 
clear,  brief  English  which  will  never  allow  misunderstandings 
to  arise  as  to  his  meaning.  The  economics  of  any  project 
form  a  large  factor,  and  so  this  engineer's  decisions  must  be 
backed  by  the  facts  and  theories  of  economics. 

Third  type, —  the  electrical  engineer  who  acts  as  liason 
engineer  between  electrical  engineering  and  other  fields. 
Here  belongs  the  consulting  electrical  engineer  who  is  re- 
tained by  any  company  to  advise  on  electrical  matters. 
Here  also  is  the  promoting  engineer,  who  sees  the  problem 
in  the  large  and  realizes  the  possibilities  of  harnessing  some 
waterfall,  building  a  manufacturing  plant  or  organizing  a 
communication  company.  He  must  know  the  needs  of 
the  communities  to  be  served  by  a  project,  the  various  uses 
to  which  the  power  can  be  put  and  the  probable  market 
for  the  company's  commodities.  Not  only  must  he  know 
these  things  well  but  he  must  also  be  able  to  explain  his 
ideas  in  clear  English  to  prospective  investors  and  to  prove 
to  them  the  real  value  and  soundness  of  the  project.  Such 
an  engineer  must  have  a  broad  knowledge  of  civil,  mechani- 
cal and  electrical  engineering.  As  much  of  his  responsibility 
is  of  a  financial  nature,  he  must  be  well  grounded  in  econ- 
omics and  versed  in  business  law  and  procedure.  He  does 
not  need  as  intimate  knowledge  of  the  details  of  electrical 
machinery  or  of  the  minor  points  in  electrical  theory  as  the 
first  two  types  of  engineers.  He  should,  however,  be  just 
as  sure  of  his  fundamentals,  just  as  keen  in  his  analysis 
and  just  as  rigorous  in  his  thinking. 

It  will  be  noted  that  all  three  engineers  described  above 
have  certain  common  requirements  in  their  education.  They 
must  all  be  familiar  with  the  fundamental  laws  of  physics, 
chemistry  and  mathematics,  and  must  know  thoroughly  the 
general  principles  of  electrical  theory  and  practice  which 
are  the  basis  of  all  electrical  engineering. 

This  text  assumes  a  knowledge  on  the  part  of  the  student 
of  the  common  facts  and  principles  of  physics,  chemistry 


12        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

and  mathematics,  and  endeavors  to  present  the  general 
principles  of  electrical  theory  in  such  a  manner  as  to  afford 
a  foundation  upon  which  the  student  of  electrical  engineering, 
regardless  of  which  branch  he  may  select,  can  build  his 
higher  technical  courses. 


SUMMARY   OF   CHAPTER  I 

THE  MODERN  TENDENCY  is  toward  greater  and  more 
efficient  utilization  of  natural  sources  of  power.  This  tendency 
is  taking  the  direction  of  larger  and  larger  central  electric 
power  stations,  the  application  of  electric  power  to  more  diverse 
uses  and  greater  refinement  of  electric  appliances.  The  field 
of  the  electrical  engineer  is  thus  gaining  in  extent  and  diversity. 
THREE  TYPES  OF  ELECTRICAL  ENGINEERS  are  needed 
in  this  development : 

First,  engineers  to  design  and  manufacture  electric  ap- 
paratus or  machinery, 

Second,  engineers  to  put  this  machinery  to  use, 
Third,   engineers  to   connect   electrical   engineering  and 

other  fields. 

CERTAIN  FUNDAMENTAL  EDUCATIONAL  REQUIRE- 
MENTS are  common  to  all  three  types,  chief  among  which 
are  the  basic  laws  of  mathematics,  physics  and  chemistry 
and  the  general  principles  of  electrical  engineering.  . 


13 


PROBLEMS   ON   CHAPTER  I 

Prob.  1-1.  If  it  requires  2.2  pounds  of  coal  per  hour  to 
produce  one  horse  power  in  a  good  modern  steam  power  plant, 
how  many  tons  of  coal  per  day  of  15  hours  are  used  by  a  power 
plant  delivering  25,000  horse  power? 

Prob.  2-1.  A  certain  oil  well  flows  2000  barrels  of  oil  per 
day.  If  this  is  burned  under  a  boiler,  how  many  horse  power 
will  it  develop  continuously?  Assume  that  1  pound  of  oil 
contains  18,000  B.t.u.  of  which  15  percent  is  available  by  this 
method  of  using  the  oil. 

Prob.  3-1.  If  it  requires  9  barrels  of  crude  oil  per  day  of 
10  hours  to  run  a  250-kilowatt  plant  at  rated  load,  what  per- 
cent of  the  energy  in  the  fuel  is  available  by  the  method  used? 

Prob.  4-1.  The  Big  Creek  reservoir  of  the  Pacific  Light 
and  Power  Company  is  4.5  miles  long,  J  mile  wide  and  has  an 
average  depth  of  34  feet.  The  effective  height  of  the  reser- 
voir above  the  water  wheel  is  1900  feet.  How  many  foot- 
pounds of  energy  are  stored  in  this  reservoir? 

Prob.  5-1.  To  how  many  tons  of  coal  averaging  14,000 
B.t.u.  per  pound  is  the  water  in  the  reservoir  of  Prob.  4-1 
equivalent  from  the  energy  standpoint? 

Prob.  6-1.  The  power  plants  in  connection  with  the  reser- 
voir of  Prob.  4-1  contain  6  water  wheels  of  20,000  horse  power 
each.  How  many  days  would  the  water  in  the  reservoir  alone 
operate  these  wheels,  assuming  that  the  average  load  is  one- 
half  the  capacity  of  the  plants  and  that  the  efficiency  at  this 
load  is  80  per  cent? 

Prob.  7-1.  The  highest  recorded  efficiency  for  water  tur- 
bines was  attained  by  the  four  6000-horse-power  wheels  at 
New  River,  Va.  Under  a  head  of  49  feet,  an  efficiency  of  93.7 
percent  was  secured.  What  flow  of  water  was  necessary  under 
these  conditions? 

Prob.  8-1.  In  the  Mississippi  River  hydro-electric  develop- 
ment at  Keokuk,  Iowa,  there  are  15  turbines  each  having  a 

14 


THE  ELECTRICAL  ENGINEER         15 

normal  rating  of  10,000  horse  power  based  on  a  head  of  32  feet. 
Under  these  conditions  they  operate  at  an  efficiency  of  about 
88  percent.  What  is  the  flow  of  water  through  them? 

Prob.  9-1.  At  full  load  the  generators  attached  to  the  tur- 
bines in  Prob.  8-1  have  a  guaranteed  efficiency  of  96.3  percent. 
How  many  kilowatts  can  each  generator  deliver  under  these 
conditions?  Data  from  General  Electric  Review. 

Prob.  10-1.  In  the  Gatun  hydro-electric  development  there 
are  3  Pelton-Francis  turbines,  each  having  a  capacity  of 
3600  horse  power  when  operating  under  an  effective  head  of 
75  feet.  The  total  flow  of  water  through  the  pen-stocks  is 
90,000  cubic  feet  per  minute.  What  is  the  efficiency  of  the 
turbines  under  these  conditions? 

Prob.  11-1.  Each  generator  attached  to  the  turbines  in  the 
Gatun  plant  has  a  guaranteed  efficiency  of  95.1  percent  when 
delivering  2000  kw.  What  horse  power  must  each  turbine 
develop  under  these  conditions?  Data  from  General  Electric 
Review. 

Prob.  12-1.  Assuming  an  efficiency  of  83  percent  for  the 
turbines  in  Prob.  11-1,  how  much  water  per  minute  must  be 
supplied  to  each  machine  at  an  effective  head  of  75  feet? 

Prob.  13-1.  Assume  the  following  conditions  in  a  good 
gas-producing  plant: 

The  producer  delivers  75  percent  of  the  energy  in  the  coal 
to  the  gas  engine. 

The  gas  engine  converts  35  percent  of  this  energy  into  me- 
chanical energy  of  the  piston. 

The  piston  delivers  90  percent  of  this  energy  to  the  shaft. 

What  is  the  over-all  efficiency  of  the  gas-producing  plant? 

Prob.  14-1.  At  $2.50  per  ton  for  coal  averaging  14,000 
B.t.u.,  what  will  it  cost  for  fuel  per  year  of  3000  hr.  to  operate 
a  100-kilowatt  electric  generator  with  a  gas-producing  engine? 
The  generator  has  an  efficiency  of  90  percent  and  the  producer 
plant  data  as  in  Prob.  13-1. 

Prob.  16-1.  A  70-horse-power  Diesel  engine  showed  on  test 
that  it  delivered  41.7  percent  of  the  energy  in  the  oil  to  the 
piston.  The  efficiency  of  the  engine  from  the  piston  to  the 
pulley  was  90  percent.  At  2  cents  per  gallon  (7.6  pounds) 
for  oil,  how  much  more  or  less  expensive  per  year  of  3000  hours 
would  it  be  to  use  a  70-horse-power  electric  motor  of  80  per 


16        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

cent  efficiency  instead  of  the  Diesel  engine?  Electricity  costs 
4  cents  per  hour  for  each  kilowatt.  (This  is  not  fair  to  the 
motor,  as  the  oil  engine  would  probably  require  much  more 
attention.)  1  pound  of  oil  =  14,500  B.t.u. 

Prob.  16-1.  A  test  was  made  on  the  shafts  and  belting  of  a 
certain  machine  shop  eight  stories  high.  A  jack  shaft  on  each 
floor  was  connected  by  belts  to  the  engine  shaft.  When  the 
shop  was  running  at  full  load,  the  sum  of  the  power  being  de- 
livered by  the  several  jack  shafts  was  196.7  horse  power.  The 
engine  was  delivering  257.2  horse  power. 

(a)  What  horse  power  was  lost  in  the  jack  shafts  and  belting? 

(6)  What  was  the  efficiency  of  the  jack  shafts  and  belting? 


CHAPTER   II 
ELECTRIC   UNITS   AND   ELECTRIC   CIRCUITS 

The  electrical  engineer  deals  primarily  with  the  genera- 
tion, transmission  and  utilization  of  electrical  power.  There- 
fore, while  occasionally  he  has  to  deal  with  electricity  at 
rest  (static  electricity),  his  usual  concern  is  with  electricity 
in  motion,  that  is,  with  electric  currents. 

8.  The  Electron  Theory.  In  every  substance  there  are 
a  large  number  of  small  particles  of  electricity.  These  are 
all  of  the  same  size  and  they  are  extremely  small.  We  call 
these  particles  electrons.*  The  theory  of  their  behavior 
under  various  conditions  is  called  the  electron  theory  and 
is  in  general  use  today. 

The  effects  produced  by  electrons  at  rest  are  called  static 
effects.  More  important  are  the  effects  produced  when 
large  numbers  of  electrons  are  in  motion. 

In  many  substances,  such  as  glass  for  instance,  the  elec- 
trons are  attached  securely  to  the  atoms  of  the  material  and 
can  be  broken  loose  only  with  great  difficulty.  These  sub- 
stances are  insulators.  In  metals  there  are  large  numbers 
of  free  electrons,  or  electrons  which  are  not  attached  and 
which  can  move  through  the  metal.  These  metals  are 
conductors.  When  the  electrons  move  we  say  that  the  metal 
conducts  a  flow  of  electricity  or  a  current  of  electricity. 

A  metal  wire  full  of  free  electrons  is  similar  to  a  pipe 
packed  full  of  sand  and  then  filled  with  water.  The  elec- 
trons may  be  likened  to  the  water  molecules  and  the  atoms 
of  the  material  to  the  grains  of  sand.  The  electrons  may 

*  See  "  The  Electron  "  by  R.  A.  Millikan  or  "  Within  the  Atom  " 
by  John  Mills. 

17 


18        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

be  made  to  filter  through  the  metal  just  as  the  water  may  be 
forced  through  the  sand.  The  stream  of  electrons  is  an 
electric  current  just  as  the  movement  of  the  water  molecules 
is  a  stream  of  water. 

Water  is  considered  an  incompressible  fluid  in  studying 
flow  in  pipes.  There  are  really  large  spaces  between  the 
water  molecules,  but  it  is  difficult  to  actually  crowd  them 
together  or  to  compress  the  water.  Anyone  who  ever  had 
an  automobile  cylinder  full  of  water  is  familiar  with  this 
fact.  We  may  similarly  treat  the  electrons  in  a  metal  as  an 
incompressible  fluid.  There  are  large  spaces  between  them 
and  yet  on  account  of  their  great  repulsion  for  one  another, 
they  are  hard  to  compress. 

Water  can  be  slightly  compressed  but  only  at  such  high 
pressures  that  we  rarely  deal  with  them.  Similarly  elec- 
tricity can  be  compressed  but  only  at  very  high  electrical 
pressures.  To  small  pressures  it  is  practically  incompres- 
sible. 

9.  The  Ampere  —  A  Unit  of  Current.  Just  as  the  flow 
of  water  in  a  pipe  is  measured  by  the  number  of  gallons 
which  pass  a  given  point  per  second,  so  the  flow  of  electricity 
along  a  conductor  is  measured  by  the  number  of  coulombs 
of  electricity  which  pass  a  given  point  per  second.  The 
coulomb  is  just  as  definite  a  quantity  of  electricity  as  the 
gallon  is  a  definite  quantity  of  water.  In  fact,  a  coulomb  is 
a  certain  number  of  electrons,  just  as  a  gallon  is  a  certain 
number  of  molecules  of  water.  There  are  about  6.3  X  1018 
electrons  in  a  coulomb.  This  number  written  out  in  full  is 
6,300,000,000,000,000,000.  It  will  readily  be  seen  why  a 
large  unit  was  chosen  for  ordinary  computations  to  avoid 
the  use  of  such  large  figures. 

An  electric  current  is  hence  usually  measured  in  "  cou- 
lombs per  second  "  rather  than  "  electrons  per  second." 
Even  the  expression  "  coulombs  per  second  "  was  considered 
too  cumbersome,  so  the  rate  of  flow  of  "  one  coulomb  per 
second  "  is  designated  as  an  ampere  rate  of  flow.  Thus  to 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        19 

designate  a  current  of  10  coulombs  per  second,  we  have 
merely  to  express  it  as  10  amperes. 

In  denning  the  ampere,  advantage  was  taken  of  the  fact 
that  if  an  electric  current  is  passed  through  a  solution 
containing  a  salt  of  a  metal,  it  will  deposit  the  metal  on  the 
negative  plate.  The  ampere  was  accordingly  legally  de- 
fined by  an  Act  of  Congress,  1894,  as  follows : 

"  The  unit  of  current  shall  be  what  is  known  as  the  inter- 
national ampere,  which  is  one-tenth  of  the  unit  of  current  of  the 
centimeter-gram-second  system  of  electromagnetic  units  and 
is  the  practical  equivalent  of  the  unvarying  current,  which  when 
passed  through  a  solution  of  nitrate  of  silver  in  water  according 
to  the  standard  specifications,  deposits  silver  at  the  rate  of  one 
thousand,  one  hundred  and  eighteen  millionths  0.00111800 
of  a  gram  per  second." 

The  ampere,  then,  is  the  common  unit  for  measuring  rate 
of  flow;  thus  a  50- watt,  110- volt  tungsten  lamp  is  said  to 
take  a  current  of  nearly  one-half  ampere  —  that  is,  a  cur- 
rent of  about  one-half  ampere  is  flowing  through  it  all  the 
time  it  is  glowing.  An  electric  flat  iron  usually  takes  about 
five  amperes,  a  one-horse-power  110- volt  motor  about  ten 
amperes,  when  running  under  full  load. 

When  a  piston-type  water  pump  is  working,  the  flow  of 
water  in  the  pipe  varies  or  pulsates.  Similarly  currents  in 
wires  are  often  variable.  In  fact>  they  ,may  alternate,  or 
change  direction  rapidly,  just  as  water  in  a  pipe  would 
alternate  in  direction  of  flow  if  the  pump  were  a  simple 
piston  without  valves. 

The  American  Institute  of  Electrical  Engineers  defines 
five  types  of  currents  thus: 

Direct  Current.  A  unidirectional  current.  In  other  words 
a  current  which  never  reverses  in  direction,  although  it  may 
vary  in  amount. 

Pulsating  Current.  A  direct  current  which  pulsates  regularly 
in  magnitude. 

Continuous  Current.  A  practically  non-pulsating  direct 
current.  Usually,  unless  otherwise  specified,  when  we  speak 


20        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

of  a  direct  current  we  mean  this  particular  kind  of  a  direct 
current. 

Alternating  Current  A  current  which  alternates  regularly 
in  direction.  Unless  otherwise  distinctly  specified,  a  periodic 
current  in  which  the  changes  in  magnitude  and  direction  are 
regularly  repeated  and  in  which  the  net  flow  in  one  direction 
along  the  wire  equals  the  net  flow  in  the  other  direction. 

With  an  alternating  current,  the  electrons  move  slightly  back 
and  forth  past  a  fixed  point  in  the  wire,  without  any  gradual 
progression  along  the  wire.  In  most  alternating-current  cir- 
cuits the  current  changes  direction  one  hundred  and  twenty 
times  a  second.  This  is  called  a  sixty-cycle  current.  The 
frequency  of  the  current  is  sixty  cycles  per  second. 

Oscillating  Current.  A  periodic  current  whose  frequency  is 
determined  by  the  circuit  constants.  The  current  that  flows 
in  the  antenna  of  a  radio-telegraph  transmitter  is  usually  an 
oscillating  current.  With  this  type  of  current  the  successive 
waves  are  often  not  of  the  same  magnitude. 

Several  types  of  currents  may  flow  in  the  same  wire 
simultaneously. 

The  laws  which  govern  the  flow  of  alternating  currents 
are  very  similar  to  those  which  govern  the  flow  of  continuous 
currents.  In  this  text  we  will  study  the  laws  of  the  electric 
circuit  when  continuous  currents  flow,  and  then  show  some 
of  the  additional  effects  which  must  be  taken  into  account 
when  the  current  varies.  Unless  otherwise  stated,  an  elec- 
tric current  in  this  text  should  therefore  be  taken  to  mean 
what  is  usually  known  as  a  direct  current,  although  more 
exactly  defined  above  as  a  continuous  current. 

10.  The  Volt  —  A  Unit  of  Pressure.  Just  as  it  requires 
pressure  to  cause  a  flow  of  water,  so  it  requires  pressure  to 
cause  a  flow  or  current  of  electricity.  Just  as  we  usually 
measure  water  pressure  in  pounds  per  square  inch,  so  we 
measure  electrical  pressure  or  electromotive  force  in  volts. 
When  a  lamp  is  rated  as  a  110- volt  lamp,  it  means  that  it 
requires  a  pressure  of  110  volts  across  the  terminals  of  that 
lamp  to  force  the  proper  amount  of  current  through  it. 
More  pressure,  that  is,  a  higher  voltage,  would  force  through 
more  current  and  probably  damage  the  lamp. 


ELECTRIC  UNITS  AND  ELECTRIC  CIRCUITS        21 

There  are  several  ways  in  which  we  can  produce  a  pressure 
tending  to  force  the  electrons  along  a  wire.  Small  pressures, 
or  voltages,  can  be  obtained  chemically  from  batteries, 
either  primary  cells  or  storage  batteries.  Their  action  is 
exactly  the  same  but  the  storage  battery  can  be  recharged 
when  exhausted;  that  is,  its  chemicals  can  be  restored  to 
their  original  condition  by  forcing  a  current  through  in  the 
reverse  direction. 

There  are  many  other  ways  in  which  electrons  may  be 
caused  to  move  along  a  wire.  The  most  important  of  these 
is  used  in  the  electric  generator,  where  the  wire  is  moved 
rapidly  through  a  magnetic  field. 

However  the  pressure  may  be  produced,  it  is  measured 
in  volts.  One  volt  as  defined  by  Act  of  Congress  is  the 
1/1. 0183th  part  of  the  pressure  delivered  at  20°  C.  by  a  stand- 
ard chemical  cell,  called  the  Weston  cell,  consisting  of  plates 
of  cadmium  and  mercury  in  an  electrolyte  of  mercurous  sul- 
phate and  cadmium  sulphate. 

11.  The  Ohm  —  A  Unit  of  Resistance.  It  was  proved 
by  Georg  Simon  Ohm  in  a  paper  published  in  1826*  that 
for  a  given  metallic  circuit  a  definite  ratio  existed  between 
the  pressure  and  the  current  —  if  the  pressure  was  doubled, 
the  current  would  also  be  doubled,  etc.  This  ratio  of 
pressure  to  current  is  called  the  resistance  of  the  circuit  and 
is  measured  in  ohms.  Thus  a  circuit  is  said  to  have  five 
ohms  resistance  when  the  ratio  of  the  pressure  to  the  cur- 
rent is  five.  That  is,  five  volts  will  force  one  ampere  through 
it,  ten  volts  will  force  two  amperes,  etc. 

Pressure         Volts          ~, 
TT—       r  =  A—        ~  =  Ohms. 
Current       Amperes 

The  resistance  of  a  wire  to  a  flow  of  current  is  similar  to 
the  frictional  resistance  of  a  pipe  to  the  flow  of  water.  The 
flow  of  electricity  can,  however,  be  measured  and  calcu- 
lated today  much  more  accurately  than  can  the  flow  of 
water. 

*  See  Friedrich  Mann's,  "  Georg  Simon  Ohm." 


22        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

12.  Ohm's  Law.  The  law  that  the  ratio  of  the  pressure 
to  the  current  in  a  given  circuit  is  constant  is  called  Ohm's 
Law,  and  is  the  fundamental  law  of  the  flow  of  electric 
currents.  In  its  simplest  form  this  may  be  written  as  fol- 
lows 

-o    .  ,  Pressure 

Resistance  =  ~  —     —  • 
Current 

In  practical  units 

~,  Volts 

Ohms  =  -r  -- 
Amperes 

In  symbols 

fi-y.  (1) 

where  R  =  resistance  in  ohms, 

E  =  electromotive  force  (or  pressure)  in  volts, 
/  =  current  in  amperes. 

The  symbol  /  is  used  for  current  under  an  international 
agreement  which  insures  having  the  same  symbols  used  in 
electrical  work  the  world  over.  This  is  very  convenient 
indeed  in  studying  books  written  in  foreign  languages. 
The  letter  7  is  taken  from  the  French  word  for  current, 
"  intensite." 

Example  1.  When  220  volts  are  applied  to  the  field  coils  of  a 
certain  motor  it  is  found  that  a  current  of  1.8  amperes  flows 
through  the  coils.  What  is  the  resistance  of  the  coils? 


«__=  122  ohms. 

1  1  .  O 

13.  Absolute  System  of  Electrical  Units.  The  Abvolt, 
Abampere  and  Abohm.  In  electrical  work  just  as  in  me- 
chanics, we  have  two  common  systems  of  units.  The  first 
or  practical  system  is  generally  used  by  engineers,  and  is 
based  upon  the  ampere,  volt  and  ohm.  In  addition  there  is 
the  c.  g.  s.  or  absolute  system.  The  units  in  this  system  have 
very  simple  relations  to  the  fundamental  units,  the  centi- 
meter, gram  and  second.  In  much  theoretical  work  it  is 
more  convenient  to  use  the  absolute  system. 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        23 

The  unit  of  current  in  this  system  is  the  abampere  and 
equals  10  amperes. 

The  unit  of  pressure  is  the  abvolt  and  equals  10~8  volt. 

The  unit  of  resistance  is  the  abohm  and  equals  10~9  ohm, 
that  is,  it  is  the  resistance  through  which  a  pressure  of  one 
abvolt  will  force  a  current  of  one  abampere. 

Prob.  1-2.  A  dry  cell  in  good  condition  has  an  internal 
resistance  of  approximately  0.08  ohm  and  an  electromotive 
force  of  1.4  volts.  If  a  wire  of  negligible  resistance  is  put 
across  the  terminals  of  a  dry  cell  of  0.08  ohm  resistance  and  1.41 
volts  e.m.f.,  what  current  would  momentarily  now  in  the  wire? 

Prob.  2-2.  State  the  resistance,  current  and  pressure  of 
Prob.  1-2  in  abohms,  abamperes  and  abvolts. 

14.  General  Application  of  Ohm's  Law.  Ohm's  law  may 
be  applied  to  an  entire  electric  circuit  or  to  any  part  of  a 
circuit. 

When  it  is  applied  to  the  entire  circuit,  care  must  be  taken 
to  make  sure  that  all  the  electromotive  force  of  the  circuit 
is  used  for  the  value  of  E,  that  all  the  current  of  the  circuit 
is  used  for  the  value  of  7  and  all  the  resistance  of  the  circuit 
for  the  value  of  R. 

Similarly  in  applying  the  law  to  only  a  part  of  the  circuit, 
care  must  be  taken  that  the  values  for  E,  I  and  R  include 
only  the  voltage,  current  and  resistance  of  that  particular  part 
of  the  circuit  under  consideration. 


10  Ohms 


20hma 


B 

FIG.  4.    The  pressure  of  100  volts  is  applied  to  the  whole  circuit. 

Example  2.     In  Fig.  4  the  generator  G  generates  an  electro- 
motive force  of  100  volts. 


24        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

(a)  How  much  current  flows  in  the  circuit? 

(6)  What  is  the  voltage  across  the  10-ohm  resistance? 

(a)  To  find  the  current  of  the  whole  circuit,  use  the  voltage 
of  the  whole  circuit  (the  100  volts  e.m.f.)  and  the  resistance  of 
the  whole  circuit  (2  +  10  +  20  +  8  =  40  ohms). 

pi       i  f\f\ 
Thus  /  =-f>=  1  T"  =  2.5  amperes. 

xC  ^Ew 

(6)  To  find  the  voltage  across  the  10-ohm  resistor,  use  the 
current  through  the  resistor  and  the  resistance  of  the  resistor. 
It  is  known  from  experiment  that  the  same  current  flows  in 
every  part  of  such  a  circuit  as  that  of  Fig.  4.  Thus  there  will 
be  2.5  amperes  flowing  through  the  10-ohm  resistor.  The 
voltage  across  the  10-ohm  resistor  must  equal  the  product  of  the 
current  through  the  resistor  and  the  resistance  of  the  resistor  or 

2.5  X  10  =  25  volts. 


Ohm's  law  should  be  very  familiar  to  every  engineer  in  all 
three  of  its  possible  forms. 


Prob.  3-2.  Assume  that  resistance  X,  Fig.  4,  is  not  20 
ohms  as  marked  but  has  an  unknown  value.  The  voltage  across 
the  resistor  A  is  measured  and  found  to  be  15  volts.  The  rest 
of  the  data  is  as  in  Fig.  4.  Compute: 

(a)  The  voltage  across  B, 

(6)  The  voltage  across  X, 

(c)   The  resistance  of  X. 

Prob.  4-2.  A  series  tungsten  lamp  is  rated  to  take  6.6  am- 
peres when  a  voltage  of  48  volts  is  applied  to  its  terminals.  It 
is  desired  to  use  two  such  lamps  in  series  on  a  115-  volt  system. 
What  resistance  must  be  connected  in  series  with  the  lamps 
to  limit  the  current  to  its  rated  value? 

Prob.  5-2.  The  voltage  at  the  generator  end  of  a  two-wire 
transmission  line  is  maintained  at  115  volts.  When  the  line 
is  carrying  42  amperes  the  voltage  at  the  receiving  end  is  112 
volts. 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        25 

(a)  What  is  the  resistance  of  the  line? 

(b)  What  is  the  voltage  at  the  receiving  end  when  the  line 
is  carrying  84  amperes? 

n.5.  Kirchhoff's  Laws.  First  Law.  Two  fundamental 
simple  laws  of  the  electric  circuit  have  received  the  name 
of  Kirchhoff  s  laws.  They  are  fundamental  natural  laws 
applied  to  the  conditions  for  the  flow  of  electric  currents  in 
a  circuit  or  network  of  conductors. 

First  Law.  Experiments  show  that  with  direct  current 
the  amount  of  current  flowing  away  from  a  point  in  a  circuit 
is  equal  to  the  amount  flowing  to  that  point. 

This  is  simply  an  expression  for  the  law  for  the  conserva- 
tion of  matter.  When  electric  currents  flow  in  a  circuit, 
no  electrons  are  lost  in  the  process.  At  any  point  in  the 
circuit  just  as  many  electrons  arrive  per  second  as  leave. 
Of  course  if  the  insulation  is  poor  and  current  leaks  off 
the  wire,  this  leakage  must  also  be  considered.  With 
moderate  voltages  the  amount  leaking  in  this  way  is  usually 
entirely  negligible. 


FIG.  5.    The  current  7  flowing  to  the  point  X  must  equal  the  sum  of 
the  currents  Ia,  ID  and  Ic  flowing  away  from  X. 

Thus  in  Fig.  5,  the  current  I  flowing  to  the  point  X  must 
exactly  equal  the  sum  of  the  currents  Ia,  h  and  Ic  which  are 
flowing  away  from  the  points.  We  express  this  by  the 

equation 

/  =  la  +  lb  +  Jc. 

In  other  words,  electricity  does  not  pile  up  or  accumulate 
at  any  point  in  an  electric  circuit,  at  least  when  a  direct 
current  is  flowing. 


2  r /-^-  '/  y      jTtid^  '* 

26        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

When  we  come  to  the  study  of  varying  currents  and 
capacitances,  we  shall  see  how  to  extend  this  law  to  cover 
alternating  or  oscillating  currents  also. 


FIG.  6.    The  water  current  7  must  equal  the  sum  of  the  currents 
/a,  /&  and  /c. 

This  law  is  exactly  in  accord  with  the  hydraulic  analogy 
of  Fig.  6.  The  current  of  water  7  flowing  to  the  point  X 
in  the  pipe  OX  must  exactly  equal  the  sum  of  the  currents 
flowing  away  from  the  point  X  in  the  three  pipes  A,  B, 
and  C,  or 

1=   Ia  +  h  +  Ic. 

16.  KirchhoiFs  Laws.  Second  Law.  The  difference  of 
the  electrical  potential  between  any  two  points  is  the  same 
regardless  of  the  path  along  which  it  is  measured.  In 
other  words,  if  there  is  an  electric  pressure  acting  between 
two  points  in  a  circuit  it  acts  with  equal  force  on  all  paths 
connecting  the  two  points.  By  difference  of  electrical 
potential  is  meant  the  electrical  pressure  or  the  voltage 
between  the  points. 

The  engineer  uses  many  terms  for  electrical  pressure 
such  as  difference  of  potential,  drop  in  potential,  electro- 
motive force,  voltage  and  so  on.  These  all  mean  electric 
pressures,  but  we  usually  speak  of  the  electromotive  force 
as  the  voltage  supplied  to  the  circuit  by  a  battery  or 
generator,  and  the  pressure .  existing  between  the  ends  of 
a  resistor  is  usually  spoken  of  as  a  drop  in  potential,  or 
simply  as  an  IR  drop. 


ELECTRIC  UNITS  AND  ELECTRIC  CIQCyiTS 

Example  3.     In  Fig.  7  there  are  three  paths  from  R 
(1)  Through  the  resistor  L, 

27 
to  S: 

/£   <=r  «/y 

(2)  Through  the  resistor  M, 
(3)  Through  the  generator  G. 
msider  the  voltage  between  the  ooints  R  and  S. 

' 

A.            1  Amperes 

,    '    R      A 

AA 

5  Ohms 

3 

/ 

2  Ohms 
I  3  Ohms) 

CO 

"  \ 

V    t 

L 

40  Ohms 

M 
20  Ohms 

1 

w 

5  Ohms 

/\       " 

B                 7  Amperes 

4  v 

Y 

FIG.  7.    The  voltage  drop  from  R  to  £  is  the  same]jwhether  com- 
puted along  the  path  RABS,  RLS  or  RXYS. 

1st,  As  computed  from  the  current  and  resistance  through 
the  40-ohm  resistor  (L) ; 

2nd,  As  computed  from  resistance  of  path  through  the  20- 
ohm  resistor  (M). 

(1)  Voltage  between  R  and  S  =  Current  through  L  times 
Resistance  of  L,  =  IL  RL, 

=  3  X  40  =  120  volts. 

(2)  Voltage  between  R  and  S  =  Current  through  path  of 
M  times  resistance  of  path  of  M, 

=  4(20  +5+5)  =120  volts. 

Thus  the  voltage,  or  potential  difference,  between  the  points 
R  and  S  is  120  volts,  whether  computed  along  the  path  of  M 
or  of  L. 

It  remains  to  compute  the  potential  difference  between  R 
and  S  along  the  path  through  the  generator  G. 

17.  Electromotive  Force  and  IR  Drop.  The  product 
of  the  current  times  the  resistance  is  called  the  IR  drop 
in  potential.  Since  the  electromotive  force  is  the  force 
which  sends  the  current  around^the  circuit,  the  IR  drop  may 
be  thought  of  as  the  force  which  opposes  the  electromotive 
force  in  sending  a  current  in  thi&  direction.  When  a  steady 


28        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

current  is  flowing,  there  is  a  condition  of  equilibrium  and  the 
opposing  forces  must  be  equal.  Therefore  the  algebraic 
sum  of  the  IR  drops  in  any  circuit  must  equal  the  algebraic 
sum  of  the  electromotive  forces. 

In  Fig.  7,  if  we  represent  the  electromotive  force  of  169  volts 
as  a  positive  force  in  the  direction  of  current  flowing  around  the 
circuit  ARSB,  we  must  represent  the  IR  drops  as  positive 
forces  in  the  direction  opposite  to  current  flow  and  the  sum  of 
the  IR  drops  must  equal  the  sum  of  the  electromotive  forces  or 
169  volts. 

The  IR  drop  through  the  generator  =3  X  7  =21  volts. 
The  IR  drop  from  A  to  R  =2x7=  14  volts. 

The  IR  drop  from  S  to  B  =2x7  =  14  volts. 

The  IR  drop  from  R  to  S  =  x    volts. 


Total  IR  drop  around  circuit  ARSB  =  (49  +  x)  volts 

The  algebraic  sum  of  IR  drops  around  a  circuit  must  equal 
the  algebraic  sum  of  the  electromotive  forces. 
Therefore 

(49  +  x}  =  169 

x    =  169  -  49 
=  120  volts. 

There  is  the  same  potential  difference  between  the  points 
R  and  S  as  found  by  computing  it  along  the  other  two  paths. 

KirchhofTs  second  law  is  very  often  stated  in  terms  of 
the  e.m.f.  and  the  IR  drop  as  follows: 


or 

2E  -  SIR  =  O, 

where  2E  is  the  algebraic  sum  of  the  electromotive  forces 
in  a  circuit  and  27  R  is  the  algebraic  sum  of  the  IR  drops 
in  the  same  circuit. 

The  reason  that  the  algebraic  sum  must  be  used  is  because 
we  may  have  electromotive  forces  opposing  each  other,  as 
when  a  generator  charges  a  storage  battery.  Also  the  cur- 
rents in  different  parts  of  a  particular  loop  in  a  network  may 
be  in  different  directions.  In  such  a  case  the  electromotive 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        29 

force  of  the  main  source  is  called  positive.  Any  electro- 
motive forces  opposing  it  are  negative.  IR  drops  opposing 
the  main  electromotive  force  are  called  positive  and  those 
aiding  it,  negative. 

In  applying  the  law  in  this  form,  great  care  must  be  ex- 
ercised to  see  that  when  one  sign  has  been  given  to  the  elec- 
tromotive force  in  the  direction  of  current  flow,  the  opposite 
sign  is  given  to  the  IR  drop  in  the  direction  of  current  flow. 
These  laws  are  of  the  greatest  assistance  in  the  determina- 
tion of  the  current,  voltage  and  resistance  relations  in  cir- 
cuits of  the  form  of  complicated  networks. 


1.5  Ohms 


1.5  Ohms 

B  ^        D 

FIG.  8.     Compute  the  current  in  the  several  parts  of  the  line. 

Example  4.  In  a  circuit  arranged  as  in  Fig.  8,  find  the  cur- 
rents delivered  by  the  generator  G  and  storage  battery  S  re- 
spectively. 

We  will  assume  the  current  to  flow  as  follows,  having  care  to 
apply  Kirchhoff  s  first  law  of  current  flow  that  the  sum  of  the 
currents  flowing  to  any  point  must  be  equal  to  the  sum  of  the 
currents  flowing  away  from  that  point. 

11  to  flow  from  A  to  C, 

12  "         C  to  D, 
t!  -  t.  C  to  E, 

(ii  —  i*  +  iz)  "  E  through  the  10-ohm  re- 

sistor to  F, 

i,  "  F  through  the  storage  bat- 

tery to  E, 

ii  -  &  "         F  to  D, 

ii  D  to  B  and  from  B  to  A 

through  generator  G. 


30        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

The  currents  may  not  in  every  case  be  in  the  direction  in 
which  we  have  assumed  them  to  flow.  If  this  is  true,  the  re- 
sults will  show  a  minus  sign  for  those  branches  in  which  we  have 
assumed  the  wrong  direction  of  flow.  The  numerical  results 
will  be  correct,  however,  and  a  negative  sign  merely  means  that 
the  current  is  actually  flowing  in  the  direction  opposite  to  that 
marked  in  the  diagram. 

As  we  have  three  unknowns,  we  must  write  three  equations. 

—  (1)  Consider  the  circuit  ACDB: 

40  =  1.5  ii  +  20  it  +  1.5  ii  +  2  ii 

=  5  ii  -f  20  i3. 
ii  =  8  -  4  it. 

—  (2)  In  the  circuit  CESFD: 

35  =  2  i,  -  1(1,  -  it)  +  20  it  -  l(ii  -  it) 
=  2  ia  -  2  ii  +  22  it. 

Substituting  value  of  ii  from  (1)  : 

35  =  2  is  -  2(8  -  4  ia)  +  22  it. 
(3)    51  =  2  iz  +  30  it. 

^  (4)  In  circuit  EMFS: 

35  =  2  i,  +  10  (ii  -  i2  +  i3) 
=  12  i3  +  10  ii  -  10  i2. 

Substituting  value  of  ii  from  (1): 

35  =  12  i3  +  10  (8  -  4  i2)  -  10  i2. 
(5)   -  45  =  12  i3  -  50  i2. 

Combining  (3)  and  (5): 

306  =  12  t,  +  180  it 
-45  =  12  t,  -  50  i, 
230  ia  =  351 

351 

amperes. 


From  (1)  ii  =  8  -  4  x  1.526 
ii  =  1.896  amperes. 

From  (3)  is  =  Sl-30f.  „  61-30X1.626  . 

2  ^ 

Thus  the  generator  G  is  delivering  1.896  amperes  and  the  bat- 
tery S,  2.610  amperes. 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        31 

The  answers  may  be  checked  by  the  use  of  Kirchhoff's  sec- 
ond law  in  the  form,  "  The  difference  of  potential  between  any 
two  points  is  the  same  along  whatever  path  it  is  computed." 
For  instance,  there  are  four  paths  between  the  points  C  and 
D,  and  the  potential  difference  computed  along  all  four  paths 
should  be  the  same. 

(1)  Along  CPD: 

The  difference  of  potential  along  this  path  is  merely  the 
IR  drop  through  the  20-ohm  resistance  at  P  when  a  current 
of  i2  =  1.526  amperes  is  flowing  through  it. 

VCD  =  i^ft  =  1.526  X  20  =  30.52  volts. 

(2)  Along  the  path  CAGBD: 

VCD  =  EG  -  ti  (2  +  1.5  +  1.5) 
=  40-5  ii 
=  40  -  9.48  =  30.52  volts  (checks  with  (1)). 

(3)  Along  the  path  CEMFD: 

VCD  =  l(ii  -  t,)  +  10  (t  i  -  ti  +  t,)  +  l(ti  -  it) 
=  12 (ti  -  ti)  +  10  t, 
=  12(1.896  -  1.526)  +  10X2.61 
=  30.54  volts   (checks  with    (1)    to    engineering 
accuracy) . 

(4)  Along  the  path  CESFD: 

VCD  =  35  -  2  it  +2(i  -  it) 

=  35  -2  X  2.61  +  2(1.896  -  1.526) 
=  30.52  volts  (checks  with  (1)). 

The  use  of  KirchhofFs  laws  will  enable  the  engineer  to 
solve  any  direct-current  network  whatever,  no  matter  how 
complicated,  if  properly  used.  In  fact  by  extending  the 
meaning  of  these  laws  they  can  be  made  to  solve  alternating- 
or  even  oscillating-current  networks. 

In  applying  the  laws,  however,  a  great  difference  in  the 
amount  of  labor  involved  in  computation  will  result  from 
different  methods  of  going  about  the  work.  Certain  gen- 
eral rules  may  be  given  for  the  methods  of  solving  an  elec- 
trical network,  and  if  followed  they  will  help  in  attaining 
speed  and  accuracy  in  such  work. 


32        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

In  order  to  solve  a  network,  first  assign  letters  to  the 
unknown  quantities  involved;  second,  form  as  many  equa- 
tions as  there  are  unknowns;  third,  solve  these  equations 
simultaneously. 

To  avoid  loss  of  time,  proceed  as  follows: 

1.  Make  a  good  diagram  of  the  circuit. 

Write  on  this  diagram  every  constant  of  the  circuit  which 
is  known,  resistances,  voltages  or  currents.  Indicate  the 
direction  of  the  voltages  or  currents  by  arrows. 

2.  Assign  letters  to  the  unknowns. 

Call  currents  i,  voltages  e,  and  resistances  r  with  sub- 
scripts. Put  an  arrow  on  each  e  or  i  to  indicate  the  di- 
rection in  which  it  is  assumed. 

Use  enough  unknowns  to  completely  "  fix  "  the  circuit, 
and  no  more.  This  means  use  as  few  unknowns  as  possible 
and  still  be  enabled  to  write  sufficient  equations  to  solve 
the  circuit. 

In  assigning  letters  to  unknowns,  apply  Kirchhoff's  laws 
as  you  go  along,  in  order  to  keep  down  the  number  of  un- 
knowns. Thus  if  three  wires  meet  at  a  point,  and  you  have 
already  called  the  current  in  two  of  them  i\  and  iz  respec- 
tively, call  the  third  current  i\  +  4  instead  of  is. 

3.  Apply  Ohm's  or  Kirchhoff's  laws  to  parts  of  the  cir- 
cuit to  obtain  equations  between  the  unknowns. 

Write  as  many  equations  as  there  are  unknowns.  There 
will  always  be  the  possibility  of  writing  more  than  this. 
One  of  the  possible  equations  not  used  at  this  time  may  be 
saved  for  a  check. 

Use  Kirchhoff's  first  law  where  it  is  possible  as  it  is  very 
simple.  In  using  the  second  law  remember  that  the  sum  of 
voltages  and  drops  may  be  taken  about  any  closed  loop  in 
the  network.  Choose  the  simplest  ones.  Of  course  loops 
involving  unknowns  must  be  used.  Be  careful  of  signs. 

4.  Solve  the  equations  algebraically  for  all  the  unknowns. 

5.  Check.     This  is  important.     Usually  a  loop  not  already 
utilized  will  give  an  excellent  check  from  the  second  law. 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        33 

Of  course  the  solution  of  simple  networks  is  easy.  The 
networks  actually  met  with  in  practice  by  the  engineer  are 
likely  to  be  complicated.  A  study  of  the  problems  of  this 
chapter  will  be  only  a  beginning  on  actual  cases  of  practice. 


y  i  y 


(X)  CX)        Cx;  DO        OOCX3 

FIG.  9.    What  voltage  exists  across  each  car? 

Prob.  6-2.     In  Fig.  9, 

Car  No.  1  is  1  mile  from  station  and  is  taking  40  amperes. 
Car  No.  2  is  3  miles  from  station  and  is  taking  20  amperes. 
Car  No.  3  is  4  miles  from  station  and  is  taking  25  amperes. 
Trolley  wire  has  a  resistance  of  0.42  ohm  per  mile.  Track 
resistance,  0.03  ohm  per  mile. 

Find: 

(a)  Voltage  across  each  car, 
(6)  Total  voltage  loss  in  line. 


0.24  Ohm  0.3  Ohm 


0.24  Ohm  0.3  Ohm 


FIG.  10.    The  generator  G  supplies  both  group  A  and  group  B  with 

Dower. 


power 

Prob.  7-2.     In  Fig.  10,  the  voltage  across  Group  A  is  116 
volts;  the  resistance  of  each  lamp  in  Group  A  is  100  ohms. 

Find: 

(a)  Voltage  across  Group  B, 

(6)  Average  resistance  per  lamp  in  Group  B, 

(c)  Line  drop  between  Generator  and  Group  A, 

(d)  Line  drop  between  Group  A  and  Group  B, 

(e)  Current  per  lamp  in  A  and  in  B. 


34        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 
v 

Prob.  8-2.  Assume  trolley  line,  Fig.  4 la,  to  be  fed  by  two 
generators,  Gi  of  560  volts  and  G2  of  555  volts.  Two  cars  are 
on  the  line,  Car  1  taking  300  amperes,  Car  2,  200  amperes. 
The  resistances  of  the  trolley  wire  and  track  are  as  marked. 

Find  the  voltage  across  each  car. 

18.  Solution  of  Series 
and  Parallel  Combinations 
of  Resistors.  Conduc- 
tance. No  special  rules 
are  needed  for  finding  the 
resistance  of  any  combina- 

f  tion  of  series  and  parallel 
FIG.  11.    The  resistance  of  the  parallel  r         .  , 

, .     ,.       ,.  ,    ,       ,  arrangements  of  resistors, 

combination  of  resistors  can  be  found         .    * 

by    the    application  of  Ohm's  and  It   is   merely  necessary  to 
Kirchhoff's  Laws.  apply    Ohm's    and     Kir- 

chhoff's  laws. 

Example  5.  Three  resistors  are  of  8,  4  and  20  ohms  re- 
spectively and  are  arranged  in  parallel  as  in  Fig.  11.  What 
is  the  resistance  of  the  combination? 


The  current  per  volt  through  the       J. 
20-ohm  resistor  20 


=  ^  =  0.05    ampere. 


The  current  per  volt  through  the    _  1 
4-ohm  resistor  4 


.  25    ampere. 


The  current  per  volt  through  the        1       n  10C 

0    ,          VV  =  o  =  0  • 125  ampere. 

8-onm  resistor  8 

The   current   per  volt  through  the  


combination 


0.425  ampere. 


This  amounts  to  assuming  a  difference  of  potential  of  one 
volt  between  points  A  and  B.  Thus,  knowing  the  voltage 
across  the  combination  (1  volt)  and  the  current  through  the 
combination  (0.425  ampere)  we  can  find  the  resistance  of 
the  combination  by  Ohm's  law: 


0.425 


=  2.35  ohms. 


ELECTRIC  UNITS  AND  ELECTRIC  CIRCUITS       35 

Note  that  the  resistance  of  a  parallel  combination  must 
be  lower  than  the  resistance  of  any  separate  branch  of  the 
combination.  Regardless  of  how  small  is  the  resistance  of 
that  branch  having  the  smallest  resistance,  the  addition  of 
the  other  branches  between  the  same  two  points  makes  it 
easier  for  currents  to  flow  between  the  two  points  and  thus 
must  lower  the  resistance  between  the  two  points. 

The  expression  "  current  per  volt  "  is  often  called  the 
conductance  of  that  part  of  the  circuit  to  which  it  is  applied. 
Note  that  the  conductance,  being  the  current  per  volt, 

equals  ^.  which  is  the  reciprocal  of  the  expression  for  the 
& 

pi 

resistance,    ^.     For  this  reason  the  conductance  is  often 

measured  in  mhos.  The  conductance  of  an  electrical  ap- 
pliance measured  in  mhos  is  the 
reciprocal  of  its  resistance  meas- 
ured in  ohms.  Thus  a  lamp  with 
a  resistance  of  10  ohms  would 
have  a  conductance  of  -  mho. 


Prob.  9-2.  The  numbers  on  the 
lines  of  Fig.  12  indicate  resistance 
in  ohms.  Compute  the  resistance 
between  A 

(a)  A  and  B,  FlG-  12-    Compute    the    re- 

(6)   B  and  C,  sistances      between      two 

(c)  A  and  C.  points  of  the  triangle. 

19.  Special  Networks,  the  Delta  and  the  Star.  A  net- 
work arranged  as  in  Fig.  13  is  called  a  Delta  or  Triangle. 
The  arrangement  of  Fig.  14  is  called  a  Star  or  Y.  These 
arrangements  are  often  combined  in  a  single  network,  as 
in  Fig.  15,  and  present  special  difficulties  in  solution.  The 
application  of  Kirchhoff's  laws  will  solve  such  a  network, 
but  to  find  the  equivalent  resistance  the  simplest  expedient 
consists  of  reducing  star  combinations  to  delta  and  vice 
versa,  as  the  occasion  requires. 


36        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

To  solve  for  the  resistance  between  points  A  and  B, 
in  Fig.  15,  for  instance,  the  star  (Rf,  R",  R'")  can  be  re- 
placed by  an  equivalent  delta  (R\t  Ri,  Rs),  and  the  arrange- 
ment of  Fig.  15  becomes  that  of  Fig.  16,  which  can  easily 


FIG.   13.    A  Delta  or  Triangle 
of  resistors. 


C  ^ 

FIG.  14.    A  Star  or  Y  arrange- 
ment of  resistors. 


m\*  z 

FIG.  15.    A  combination  of  Star  FIG.  16.     This  arrangement   of 

and    Delta    arrangements    of  resistors  is  delta  equivalent  of 

resistors.  the  arrangement  of  Fig.  15. 

be  solved  for  the  resistance  between  A  and  B.  It  can  be 
proved  that  star  (R'}  R",  R'")  is  the  equivalent  of  delta 
CRi,  #2,  #3)  if  Ri  is  made  equal  to 

R'R"  +  R"R"'  +  R'R1" 


and  Rz  to 
and  Rs  to 


w 

R'R"  +  R"R'"  +  R'R1 

R" 
R'R"  +  R"R'"  +  R'R 

R'" 


nt 


ELECTRIC  UNITS  AND  ELECTRIC  CIRCUITS       37 

Knowing  R',  R"  and  R'",  the  values  of  the  resistances 
of  the  delta  Ri,  R%  and  #3  can  thus  be  determined.  Of  course 
X  and  Z  in  Fig.  16  must  also  be  known  if  the  resistance 
between  A  and  B  is  to  be  found. 

Note:  Special  care  must  be  taken  in  the  notation.  In 
the  equivalent  delta  arrangement,  Ri,  R%  and  R3  must  be  op- 
posite in  position  with  respect  to  R' ,  R"  and  R'"  of  the  star. 

Prob.  10-2.  Prove  that  the  delta  arrangement  of  Fig.  13 
is  equivalent  to  the  star  arrangement  of  Fig.  14  if 


and 


R' 
R'R"  +  R"R'"  +  R'R'" 

R" 
R'R"  +  R"R'"  +  R'R'" 

R'" 


Prob.  11-2.      Prove  that  the  star  arrangement  of  Fig.  14  is 
equivalent  to  the  delta  arrangement  of  Fig.  13  if 

R'     = 


/ii  +  #2  +  Ra 

R"  =/p  jy'+g 

iti  ~r  /t/2  ~t~  •fl'3 
£>,#„ 

and  R'"  = 


Ri  +  R%  +  ^3 
Hint :    Use  equations  of  conductances. 

Prob.  12-2.     Compute   the   resistance   between   the    points 
A  and  B,  Fig.  15,  where 

R'  =5  ohms, 
R"  =  7  ohms, 
R "'  =  9  ohms, 
X  =4  ohms, 
Z  =8  ohms. 

20.    Measurement  of  Current,  Voltage  and  Resistance.* 

The  commercial  method  of  measuring  current  is  by  means 

*  For  details  of  construction,  calibration  and  use  of  electrical  instru- 
ments, see  Laws'  "Electrical  Measurements." 


38        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

of  an  ammeter  inserted  in  the  circuit  at  the  point  where  it 
is  desired  to  know  the  current. 

In  Fig.  17,  ammeter  A  measures  the  current  in  the  line 
AB;  ammeter  A\  measures  the  current  in  branch  BE  or 
the  current  taken  by  the  lamp  LI.  Since  the  same  current 
must  flow  away  from  Li  as  flows  to  it,  the  ammeter  could 
just  as  properly  be  placed  between  LI  and  the  point  E. 


FIG.  17.    The  reading  of  the  ammeter  A  is  the  sum  of  the  readings 
of  ammeters  AI  and  A2. 

Ammeter  A  2  measures  the  current  taken  by  lamp  L2. 
Ammeter  A2  could  be  placed  anywhere  in  the  line  BCDE 
to  measure  this  current.  The  sum  of  the  readings  of  A\ 
and  A 2  should  equal  that  of  A. 

An  ammeter  must  have  a  very  low  resistance  in  order 
that  it  may  not  appreciably  increase  the  resistance  of  the 
line  in  which  it  is  inserted.  The  potential  difference  across 
the  terminals  of  a  well-known  make  of  these  instruments 
is  about  0.05  volt  for  full-scale  reading.  Thus  a  10-ampere 

ammeter  would  have     '        or  0.005  ohm  resistance,  while  a 

100-ampere  instrument  would  have     '„„     =  0 . 0005  ohm. 

luu 

The  potential  difference  between  two  points  is  commer- 
cially measured  by  means  of  a  voltmeter,  one  terminal  of 
which  is  attached  to  one  point  and  the  other  terminal  to 
the  other  point.  Thus  in  Fig.  17a,  voltmeter  V\  measures 
the  difference  of  potential  or  voltage  between  the  terminals 
of  the  lamp  LI.  Voltmeter  V%  measures  the  potential 


ELECTRIC  UNITS  AND  ELECTRIC  CIRCUITS       39 

difference  or  voltage  between  the  terminals  of  lamp  L2. 
A  voltmeter  F3  attached  to  A  and  B  would  measure  the 
potential  difference  between  the  points  A  and  B.  This 
value  might  differ  slightly  from  .the  voltage  across  the 
lamp  Li,  because  the  voltage  across  AB  equals  the  voltage 
across  the  lamp  LI  plus  the  drop  in  that  amount  of  wire 
which  is  used  to  connect  the  lamp  to  the  points  A  and  B. 


B  D 

FIG.  17a.     The  voltmeter  Fi  measures  the  voltage  drop  across  the 
lamp  I/i.    Voltmeter  F2  measures  the  drop  across  the  lamp  L2. 

A  voltmeter  should  have  a  very  high  resistance  in  order 
that  it  may  not  draw  an  appreciable  current  from  the  cir- 
cuit to  which  it  is  connected.  The  resistance  of  a  150-volt 
instrument  is  in  the  neighborhood  of  20,000  ohms  or  some- 
what more  than  100  ohms  per  volt. 

For  accurate  work  we  must  remember  that  when  am- 
meters and  voltmeters  are  connected  into  a  circuit  they 
alter  the  circuit  conditions  by  adding  new  resistances  and 
branches. 

Thus  a  voltmeter  connected  between  C  and  D  of  Fig.  17 
will  measure  the  voltage  drop  in  the  lamp  plus  the  voltage 
drop  in  the  ammeter.  To  obtain  the  actual  voltage  across 
the  lamp  we  should  subtract  the  computed  drop  in  the 
ammeter  from  the  voltmeter  reading.  £. 

If  the  voltmeter  is  connected  between  ^  and  D  it  will 
then  read  the  correct  voltage  across  the  lamp,  but  in  this 
case  the  ammeter  will  read  not  only  the  lamp  current  but 
also  the  voltmeter  current.  The  latter  may  be  computed 
and  subtracted  to  obtain  the  actual  lamp  current, 


40       PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


In  measurements  taken  on  a  circuit  carrying  fairly  large 
currents  at  moderate  voltages,  the  drops  in  ammeters  and 
the  current  taken  by  voltmeters  are  usually  negligible. 
If  we  were  commercially  testing  a  20-horse-power,  550- volt 
motor,  for  instance,  we  should  not  ordinarily  make  allow- 
ance for  the  effect  of  putting  in  the  meters. 

21.  The  Wheatstone  Bridge.  Resistances  may  be  meas- 
ured roughly  by  using  an  ammeter  and  a  voltmeter.  The 
resistance  to  be  measured  is  connected  to  a  circuit  which 
will  pass  through  it  a  current  it  can  safely  carry.  This 

current  is  measured  with 
an  ammeter  and  the  drop 
due  to  it  with  a  volt- 
meter. Corrections  as 
outlined  in  the  last  sec- 
tion are  made  to  these 
readings.  The  quotient 
then  gives  the  desired 
value  of  resistance. 

A  more  accurate 
method  of  measuring 
resistance  is  by  some 
form  of  the  Wheatstone  bridge.  This  bridge  consists  of 
an  arrangement  of  four  resistors,  a  steady  source  of  elec- 
tromotive force  (such  as  a  battery  cell)  and  a  galvanometer. 
The  four  resistors  are  arranged  in  two  parallel  paths  as 
shown  in  Fig.  18,  where  the  resistors  R\  and  R2  form  path 
ABC,  and  resistors  Rs  and  R*  form  path  ADC.  The 
battery  is  attached  to  the  terminals  A  and  C  of  these  parallel 
paths  and  the  galvanometer  (G)  is  bridged  across  the  inter- 
mediate points  B  and  D.  When  the  resistances  of  the 
resistors  have  been  so  adjusted  that  no  current  flows  through 
the  galvanometer,  the  points  B  and  D  must  be  at  the  same 
electrical  potential. 

If  points  B  and  D  are  at  the  same  potential,  the  potential 
drop  from  A  to  D  must  equal  the  potential  drop  from  A 


FIG.  18.    The  conventional  diagram  of 
Wheatstone's  bridge. 


ELECTRIC  UNITS  AND  ELECTRIC  CIRCUITS       41 

to  B.    Similarly,  the  drop  from  D  to  C  must  equal  the  drop 
from  B  to  C. 

Let  the  current  in  branch  ABC  be  i\  and  the  current  in 
branch  ADC  be  z'2. 


Drop  A  to  B  = 
Drop  A  to  D  = 

tifli  =  *'2#3.  (1) 

Drop  B  to  C  =  iiR2, 
Drop  D  to  C  =  *'2#4, 

(2) 


T*V«     •  i       /-«\    1  /r»\  llRl  l^tliZ 

Divide  (1)  by  (2)         T--  =  T-  . 


Therefore  f^lf-  (3) 

If  any  three  of  these  resistances  are  known  in  amount, 
the  fourth  can  be  found  from  equation  (3). 

In  practice  two  of  the  resistors  are  usually  constructed 
so  that  they  may  be  made  10,  100  or  1000  ohms  in  value. 
A  third  resistance  is  variable  between  wide  ranges,  usually 
from  several  thousand  ohms  to  a  tenth  or  a  hundredth  of 
an  ohm.  The  fourth  is  the  resistor  of  which  it  is  desired  to 
measure  the  resistance. 

Thus  in  Fig.  18,  the  resistors  R  i  and  R2  might  well  have 
values  of  10,  100  or  1000  ohms,  while  R*  was  the  resistor 
with  the  wide-range  adjustment.  #3  would  be  the  same 
fraction  or  multiple  of  R*  that  Ri  was  of  Rz. 

Example  6.  Assume  that  in  Fig.  18,  Ri  was  set  at  10  ohms 
and  #2  at  1000,  and  that  R*  had  to  be  set  at  4124.6  in  order  to 
balance  the  bridge,  that  is,  to  obtain  a  condition  in  which  there 
was  no  deflection  of  the  galvanometer  needle.  What  would 
be  the  value  of  _R3? 

Ri^ 


]   of  4124.6 


1000 
=  41.246  ohms. 


42        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Prob.  13-2.  Fig.  19  represents  a  Wheatstone  bridge  in 
which  the  two  resistors  R  i  and  R2  of  Fig.  10  are  replaced  by  a 
single  wire  BD  of  high  resistance  and  uniform  cross-section. 


c/- 


FIQ.  19.    The  slidewire  Wheatstone  bridge. 

A  balance  is  obtained  by  sliding  the  contact  point  A  along  the 
wire.     Write  the  equation  for  this  form  of  bridge. 

Prob.  14-2.     Fig.  20  represents  a  Wheatstone  bridge  set  up 
"  Varley  Loop  "  test  for  the  location  of  a  ground  in  a 


for 


FIG.  20.    A  Wheatstone  bridge  arranged  for  a  "Varley  Loop"  test, 
to  locate  the  place  F,  where  the  line  is  grounded. 

cable.  The  "  return  "  wire  need  not  have  the  same  resistance 
per  foot  as  the  line.  Derive  the  equation  for  the  distance  (X) 
out  on  the  line  to  the  grounded  point  "  F."  Resistance  per 
foot  of  "  line  "  is  known. 


SUMMARY   OF   CHAPTER  U 

THE  VOLT  ( =  108  abvolts)  is  the  practical  unit  of  pressure. 

THE  AMPERE  (=  KT1   abampere)   is  the  practical  unit  of 

current. 

THE  OHM  ( =  109  abohms)  is  the  practical  unit  of  resistance. 

OHM'S  LAW: 

Volts  E 

Amperes  =  — I  =  — 

Ohms  R 

Volts  =  Amperes  X  Ohms  E  =  IR 

Volts  E 

Ohms  =  - —  R  =  - 

Amperes  I 

KIRCHHOFF'S  LAWS;   used  in  the  solution  of  networks: 
First  Law :      The  sum  of  all  the  currents  flowing  to  any  point 

equals  the  sum  of  all  the  currents  flowing 

from  that  point. 
Second  Law :     The  algebraic  sum  of  the  electromotive  forces 

in  any  circuit  equals  the  sum  of  the  resistance 

drops. 

CONDUCTANCE  is  the  reciprocal  of  resistance. 
CURRENT  is  usually  measured  by  an  ammeter. 
VOLTAGE  is  usually  measured  by  a  voltmeter. 
RESISTANCE  may  be  measured  by  applying   Ohm's  law 
to  values  of  the  current  and  voltage  or  by  the  use  of  a  Wheat- 
stone  bridge. 


43 


PROBLEMS    ON   CHAPTER   II 

Prob.  16-2.     In  Fig.  21, 

Resistance  of  A  =  100  ohms, 
Resistance  of  B  =  120  ohms, 
Resistance  of  C  =  160  ohms. 


-110  Volts- 


FIG.  21.    A  combination  of  series  and  parallel  arrangements  of 

resistors. 
Find: 

(a)  Current  through  each  resistance, 

(6)  Resistance  of  parallel  combination  (A  and  B), 

(c)  Combined  resistance  of  system, 

(d)  Voltage  across  each  resistance. 

n/WWVW- 


MMMMM 

FIG-  22.     A  combination  of  series  and  parallel  arrangements  of 
resistors. 

Prob.  16-2.     In  Fig.  22, 

Voltage  from  A  to  B  =  40  volts, 
Current  through  resistance  x  is  2.5  amperes, 
Resistance  y  =  5  ohms, 
Resistance  z  =  4  ohms. 
44 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        45 

Find: 

Current  through  y,     Resistance  of  x, 
Current  through  z,     Voltage  B  to  C. 

Prob.  17-2.  In  Fig.  23,  each  lamp  takes  12  amperes.  Re- 
sistance of  AB  =  EC  =  0.04  ohm.  Resistance  of  CD  =  MK 
=  0.03  ohm.  Resistance  of  EF  =  KF  =  0.02  ohm.  What  is 
the  voltage  across  Group  I  and  Group  II,  if  the  terminal  volt- 
age of  Gi  =  120  volts  and  of  G2  =  125  volts? 

+A  B  C 4D 


-M  K  F  -E 

FIG.  23.    Two  lamp  loads  fed  by  two  generators,  one  at  each  end 
of  the  line. 

Prob.  18-2.  A  storage  battery  of  240  cells  in  series  is 
"  floated  "  at  the  end  of  a  4-mile  trolley  line,  the  resistance  of 
which  (line  and  return)  is  0.08  ohm  per  mile.  The  generator 
voltage  is  550  volts;  each  cell  has  an  e.m.f.  of  2.1  volts  and  an 
internal  resistance  of  0.001  ohm.  (a)  What  current  will  the 
cells  supply  to  the  line,  when  there  are  5  cars  at  the  battery 
end  of  the  line,  each  taking  65  amperes?  (6)  What  will  be  the 
terminal  voltage  of  the  set  of  battery  cells? 

Prob.  19-2.  What  current  will  the  cells -in  Prob.  18-2  take 
when  there  is  no  load  on  the  line? 

Prob.  20-2.  What  current  will  the  generator  be  delivering 
when  the  5  cars  of  Prob.  18-2  are  half  way  between  station  and 
battery?  Cars  are  carrying  the  same  current  as  in  Prob.  18-2. 

Prob.  21-2.  What  current  will  the  battery  be  delivering  to 
or  receiving  from  the  line  in  Prob.  20-2? 

Prob.  22-2.  If  there  are  only  two  cars  on  the  line  in  Prob. 
20-2,  each  taking  75  amperes,  (a)  what  current  is  the  generator 
delivering?  (6)  What  current  is  the  battery  receiving  or  deliver- 
ing? 

Prob.  23-2.  Six  storage  cells  arranged  as  in  Fig.  24  in  two 
parallel  sets  of  three  cells  in  series  are  discharging  through  a 


46        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

resistor  of  1.2  ohms.  Each  cell  normally  has  an  e.m.f.  of  2.1 
volts  and  internal  resistance  of  0.02  ohm,  but  one  cell  in  set  A 
has  "  gone  bad  "  and  has  an  e.m.f.  of  1.8  volts  and  an  internal 

resistance  of  0.14  ohm.  What 

~  current  is  supplied  to  R  by 

each  set  of  cells? 


T 


Prob.  24-2.     If  the  resist- 
ance of  R,    Fig.    24,    is    re- 
duced to  0.02  ohm,  and  the 
other   data  left  as  in  Prob. 
23-2,  what  current  will  flow 
FIG.    24.     Batteries  A  and    B    con-   through  each  set  of  ceUs  and 
nected   in  parallel  are    supplying  in  what  direction? 
current  to  the  resistor  R. 

Prob.  26-2.  If  the  resist- 
ance of  R,  Fig.  24,  is  increased  to  2.4  ohms  and  the  other  data 
left  as  in  Prob.  23-2,  what  current  will  flow  through  each  set 
of  cells  and  in  what  direction? 

Prob.  26-2.  A  building  is  supplied  by  a  two-wire  system, 
the  wiring  diagram  of  which  is  as  per  Fig.  25.  Values  on  lines 
represent  the  resistances  of  the  various  sections  of  line  wire. 


0.08 


R       0.3 


0.6 


1 

yv 

?  <*, 

q  ps 

t           a 

\y 

( 

0.08 
0.03 

-. 

0.3                    0.5 
0.04                  0.04 

-. 

0.6 
0.06 

T~~Vos 

< 

'            r<rV>            * 

0.04             \J/           0.04 

r 

ToGe 
0.05                           125Vo1 

0.6 


V  P 

FIG.  25.    The  wiring  plan  of  a  two-story  factory. 


Motor  M  takes  40  amperes.  Each  lamp  takes  4  amperes  and 
the  resistance  of  each  is  constant,  (a)  Find  the  voltage  across 
each  set  of  lamps.  (6)  Draw  a  diagram  indicating  the  amount 
and  direction  of  the  current  in  each  section  of  line. 

Prob.  27-2.  If  the  motor  in  Prob.  26-2  is  not  running,  what 
will  be  the  voltage  across  each  group  of  lamps  in  Fig.  25?  Each 
lamp  is  assumed  to  take  4  amperes. 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        47 

Prob.  28-2.  If  the  jumpers  MN  and  OP  are  left  out  of 
Fig.  25,  what  will  be  the  voltage  across  each  set  of  lamps? 
Assume  each  lamp  to  take  4  amperes  and  the  motor  40  amperes. 

Prob.  29-2.  An  electric  railroad  is  supplied  with  power  by 
substations  10  miles  apart.  The  trolley  wire  is  connected  at 
frequent  intervals  with  a  heavy  copper  conductor  called  a 
"  feeder."  The  two  may  be  considered  jointly  as  equivalent 
to  a  single  conductor  whose  resistance  is  0.18  ohm  per  mile. 
The  return  path  from  the  cars  to  the  sub-stations  consists  of 
the  track  which  is  well  bonded  at  the  joints  and  has  a  resistance 
of  0.0305  ohm  per  mile.  At  a  certain  time  two  cars  are  running 
in  the  section  between  these  sub-stations,  one  taking  200 
amperes  at  a  distance  of  4.5  miles  from  one  end,  the  other  taking 
240  amperes  at  a  distance  of  3  miles  from  the  other  end.  The 
voltage  at  the  busbars  of  each  power  station  is  700.  What  is 
the  voltage  between  trolley  and  track  at  each  car? 


O 


FIG.  26.    The  voltage  of  car  C  is  "boosted"  by  means  of  the 
"booster"  feeder  B  and  F. 

Prob.  30-2.  Fig.  26  represents  a  method  of  raising  line  voltage. 
T  is  the  trolley  wire  and  R  the  rails.  F  is  a  feeder  which  is 
connected  to  the  trolley  wire  at  a  distance  of  4.2  miles  from 
the  power  station.  B  is  a  booster,  i.e.,  a  generator  used  to 
increase  the  voltage  applied  to  the  feeder.  The  following  con- 
ditions exist.  The  terminal  voltage  of  the  generator  is  600. 
The  trolley  wire  has  a  resistance  of  0.064  ohm  per  1000  feet. 
The  rails  form  a  return  conductor  whose  resistance  is  0.0061 
ohm  per  1000  feet.  The  booster  armature  has  a  resistance  of 
0.024  ohm  and  generates  an  e.m.f.  of  100  volts.  The  total 
resistance  of  the  feeder  is  0.82  ohm.  C  is  a  train  requiring  a 
current  of  500  amperes.  What  is  the  lowest  voltage  which  the 
train  can  receive  in  the  region  between  the  power  station 
and  the  point  of  connection  between  the  feeder  and  trolley? 


48        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Suggestion:     Let  x  be  the  distance  from  the  power  station  to 
the  train.     The  minimum  voltage  occurs  when  dv/dx  is  zero. 


31-2.  A  storage  battery  and  a  motor  armature  are 
connected  through  leads  of  0.5  ohm  resistance  to  form  a  closed 
circuit.  The  chemical  e.m.f.  of  the  battery  and  its  internal 
resistance  are  120  volts  and  0.12  ohm  respectively.  If  the 
motor  armature  has  a  resistance  of  0.45  ohm,  what  e.m.f.  must 
it  generate  by  rotation  in  order  that  the  battery  may  discharge 
at  its  normal  rate  of  10  amperes? 

Prob.  32-2.  Two  storage  batteries  are  connected  in  parallel 
to  be  charged  from  110-  volt  mains.  As  110  volts  is  not  suffi- 
cient to  charge  the  batteries  at  the  desired  rate,  a  booster  is 
placed  in  series  with  one  of  the  mains,  connected  so  that  its 
voltage  adds  to  the  line  voltage.  The  e.m.f.  induced  in  the 
booster  armature  is  6  volts  and  the  armature  resistance  is  0.06 
ohm.  The  internal  resistances  of  the  batteries  are  0.20  and  0.33 
ohm  respectively.  The  e.m.f.'s  induced  by  chemical  action 
in  the  batteries  are  108  volts  and  110  volts  respectively.  Com- 
pute the  current  in  each  part  of  the  circuit. 


0.005  Ohm 


H-r 


0.005  Ohm 


FIG.  27.     The  voltage  of  the  "load"  is  made  more  nearly  constant 
by  means  of  the  storage  battery  B  and  the  booster  D. 

Prob.  33-2.  The  power  received  at  A  B,  Fig.  27,  is  used  to 
operate  a  group  of  elevators  and  fluctuates  greatly.  A  storage 
battery  (B)  is  used  to  equalize  the  load,  in  conjunction  with  a 
small  dynamo  (D)  called  a  booster.  The  field  coils  of  the 
booster  are  not  shown  and  need  not  be  considered.  The  booster 
is  regulated  in  such  a  manner  as  to  assist  the  battery  to  charge 
when  the  elevators  are  idle  and  to  assist  it  to  discharge  when  their 
demand  is  high. 

The  resistance  of  the  booster  armature  is  0.012  ohm.  The 
internal  resistance  of  the  battery  is  0.015  ohm.  In  its  average 


ELECTRIC   UNITS  AND  ELECTRIC  CIRCUITS        49 

state  of  charge  the  battery  sets  up  an  e.m.f.  of  225  volts  by 
chemical  action.  The  voltage  between  mains  at  the  primary 
source  of  power  AB  is  230  volts. 

(a)  What  must  be  the  value  and  what  the  direction  of  the 
e.m.f.  induced  in  the  booster  armature  in  order  that  the  battery 
may  be  charged  by  a  current  of  240  amperes  when  the  elevators 
are  idle? 

(6)  What  value  and  direction  must  the  booster's  e.m.f.  have 
in  order  that  the  battery  and  primary  source  may  each  supply 
half  the  current  when  the  elevators  require  450  amperes? 


FIG.  28.    The  Wheatstone  Bridge. 


Prob.  34-2.  A  schematic  diagram  of  a  Wheatstone  bridge 
is  shown  in  Fig.  28.  If  the  galvanometer  G,  whose  resistance 
is  18  ohms,  registers  no  current,  what  is  the  resistance  of  X? 

I  Prob.  35-2.  In  Fig.  28,  the  battery  has  an  e.m.f.  of  2  volts 
and  an  internal  resistance  of  0.1  ohm;  lead  and  contact  resistance 
is  negligible,  (a)  What  would  be  the  percentage  error  intro- 
duced in  determining  the  resistance  of  X  if  the  galvanometer 
failed  to  deflect  when  a  current  of  10~5  ampere  was  passing 
through  it  from  A  to  £?  (6)  Would  the  true  value  of  X  be 
higher  or  lower  than  the  value  obtained  by  this  measurement? 

Prob  36-2.  Two  wires  each  having  a  resistance  of  0.32 
ohm  connect  the  terminals  of  a  battery  to  two  points  a  and  6, 
between  which  are  three  branches  whose  resistances  are  re- 
spectively 4,  8.25  and  12.5  ohms.  The  battery  generates  by 
chemical  action  an  e.m.f.  of  124  volts  and  has  an  internal  re- 
sistance of  0.16  ohm.  (a)  What  current  flows  in  each  branch 


50        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


from  a  to  6? 
minals? 


(6)  What  is  the  voltage  across  the  battery's  ter- 


Prob.  37-2.  An  electric  circuit  consists  of  the  armature  of 
a  separately  excited  generator,  two  connecting  wires  and  a 
storage  battery,  all  in  series.  The  armature  has  a  resistance 
of  0.25  ohm  and  generates  an  e.m.f.  of  115  volts  by  its  rotation. 
Each  of  the  connecting  wires  has  a  resistance  of  0.2  ohm.  The 
battery  has  a  resistance  of  0.12  ohm  and  generates  by  chemical 
action  an  e.m.f.  of  108  volts  opposed  in  direction  to  the  e.m.f. 
of  the  generator,  (a)  Find  the  terminal  voltage  of  the  generator. 
(6)  Find  the  terminal  voltage  of  the  battery. 


FIG.  29. 


1.2  Volts 

An  insulation  test  of  a  motor. 


Prob.  38-2.  The  leakage  conductance  of  the  insulation  be- 
tween a  pair  of  wires  in  a  certain  telephone  cable  is  7  mhos  per 
mile.  The  resistance  of  the  conductor  is  r  ohms  per  mile. 
Considering  differential  elements  of  insulation  as  connected  in 
parallel  by  the  conductors,  compute  the  current  input  to  the 
power  end  of  the  line  when  a  potential  of  V0  volts  is  applied 
to  this  end.  Assume  that  the  distant  end  of  the  line  is  open. 
Compute  the  home-end  resistance  of  the  line  by  dividing  the 
input  voltage  by  the  input  current.  If 

7  =  0.0008  mho  per  mile, 
r  =  53  ohms  per  mile, 
length  of  cable  =  10  miles, 
V0  =  100  volts, 

compare  the  home-end  resistance  obtained  by  the  above  method 
with  the  leakage  resistance  obtained  by  dividing  the  leakage  re- 
sistance per  mile  by  the  length  of  the  cable.  This  problem  re- 
quires knowledge  of  hyperbolic  functions, 


ELECTRIC  UNITS  AND  ELECTRIC  CIRCUITS        51 


f"  Prob.  39-2.  In  Fig.  29,  one  brush  of  the  motor  is  attached 
to  one  side  of  a  11 5- volt  circuit.  The  other  side  of  the  circuit 
is  attached  to  the  frame  of  the  motor  through  a  voltmeter  which 
reads  1.2  volts.  The  resistance  of  the  voltmeter  is  18,650  ohms. 
What  is  the  insulation  resistance  between  the  armature  and 
the  frame  of  the  motor? 

Prob.  40-2.  Find  the  resistance 
between  the  points  A  and  B  in 
Fig.  30.  The  numbers  on  the  figure 
indicate  resistances. 

Prob.  41-2.  Compute  the  re- 
sistances in  Fig.  30  between  the 
points 

(1)  A  and  Z>, 

(2)  B  and  D, 

(3)  C  and  D. 

Prob.  42-2.  A  single-track  rail- 
road line  is  5  miles  long  and  is 
supplied  with  power  from  power 
stations  at  its  two  ends.  The  positive  conductor  consists  of 
a  third  rail  reinforced  by  a  copper  cable  in  parallel,  the  com- 
bination having  a  resistance  of  0.0493  ohm  per  mile.  The 
negative  conductor  consists  of  a  pair  of  bonded  track  rails  in 
parallel,  the  combination  having  a  resistance  of  0.0305  ohm 
per  mile.  The  station  at  one  end  maintains  a  potential 
difference  of  600  volts  and  that  at  the  other  end  575  volts 
between  positive  and  negative  conductors. 

At  what  point  along  the  line  would  an  electric  locomotive 
requiring  1200  amperes  receive  a  minimum  voltage  between 
third  rail  and  track?  What  would  this  voltage  be,  and  what 
current  would  each  station  supply  when  the  locomotive  was 
at  the  point  in  question? 


rangement  of  resistors. 


CHAPTER  III 
ELECTRIC   POWER   AND   ENERGY 

A  current  of  electricity  flowing  along  a  conductor  has  been 
likened  to  a  current  of  water  flowing  in  a  pipe.  The  analogy 
is  so  close  that  we  may  use  the  same  method  to  compute 
the  power  required  to  maintain  either  a  current  of  water 
or  a  current  of  electricity. 

22.  The  Power  Equation.  A  pump  which  is  forcing  a 
steady  current  of  7  pounds  of  water  per  second  against  a 
head  or  pressure  of  E  feet  is  doing  work  at  the  rate  of  I  X  E 
foot-pounds  per  second.  An  electric  generator  which  is 
causing  a  steady  electric  current  of  I  amperes  to  flow  under 
a  pressure  of  E  volts  is  doing  work  at  the  rate  of  7  X  E 
volt-amperes. 

Since  power  is  the  time  rate  of  doing  work,  the  pump  is 
said  to  have  a  power  of  7  X  E  foot-pounds  per  second  and 
the  generator  a  power  of  /  X  E  volt-amperes  or  watts. 
In  direct-current  circuits 

1  volt-ampere  =  1  watt. 

In  the  form  of  an  equation  this  is  usually  expressed 
P  =  /  X  E, 

in  which 

P  =  power  in  watts, 
I  =  current  in  amperes, 
E  =  pressure  in  volts. 

For  convenience  1000  watts  is  called  a  kilowatt. 

The  power  equation,  as  well  as  Ohm's  and  Kirchhoff's 
laws,  applies  in  simple  form  only  to  direct-current  circuits. 
It  can  also  be  extended  to  cover  the  power  involved  in  the 
flow  of  alternating  currents. 

52 


ELECTRIC  POWER  AND  ENERGY  53 

Example  1.  At  what  rate  is  a  pump  working  which  is 
delivering  3000  pounds  of  water  per  second  against  a  head  of 
22  feet? 

P  =  IE 

=  3000  X  22  ft.  Ib.  per  sec. 

66,000 
or  -T^TT-   =  120  h.p. 


Example  2.     A   generator  is   delivering   300  amperes  at  a 
pressure  of  220  volts.     What  power  is  it  delivering? 

P  =  IE 

=  300  X  220 
=  66,000  watts 


or  -  =  66  kilowatts. 

JLjUUu 

The   following    relations   exist    between    electric   units  of 
power  and  mechanical  units  of  power: 

1  watt  =  0.737  foot-pound  per  second 

1  kilowatt  =1.34  horse  power 

1  foot-pound  =  1.356  watt-seconds 

1  horse  power  =  746  watts 

or  J  kilowatt  (practically). 

23.  Voltage  Not  a  Force.     In  the  case  of  the  flow  of  a 
fluid  such  as  water,  the  equation  for  the  power  is 

Power  =  Pressure  X  Current. 
Therefore 

Power 


Pressure  = 


Current 


The  thing  we  call  pressure  is  therefore  merely  the  power 
per  unit  current.  Since  the  same  equations  hold  in  electric 
power,  the  voltage  also  must  be  the  power  per  unit  current 
of  electricity.  Because  of  the  similarity  of  the  equations 
for  water  power  and  for  electric  power,  the  electric  voltage 
is  called  the  electric  pressure.  The  similarity  between  elec- 
tric voltage  and  water  pressure  does  not  extend  much  fur- 
ther, for  while  water  pressure  can  be  expressed  in  terms  of 
mechanical  force,  electric  voltage  remains  strictly  the 


54       PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

power  per  unit  current  and  is  not  a  force  in  the  usual  sense 
of  the  word.  There  is  such  a  quantity  as  an  electric  force 
which  is  explained  in  Chapter  XIII. 

24.  Use  of  the  Power  Equation.  The  same  precautions 
must  be  exercised  in  the  use  of  the  power  equation  (P  =  IE) 
that  are  exercised  in  the  use  of  Ohm's  law.  The  power 
equation  like  Ohm's  law  may  be  applied  to  the  whole  of  an 
electric  current  or  to  any  part.  But  also,  as  in  the  use  of 
Ohm's  law,  when  the  power  of  the  whole  circuit  is  to  be 
computed,  the  summation  of  the  electromotive  forces  of  the 
entire  circuit  must  be  used  for  E.  Similarly  when  the 
power  of  only  a  part  of  a  circuit  is  to  be  computed,  the  value 
used  for  E  must  be  the  potential  difference  across  that  part 
of  the  circuit  only,  and  the  value  of  the  current  used  for  / 
must  be  the  current  through  the  same  part  of  the  circuit  only. 

2  Ohms 


5  Amperes 


2  Ohms 


FIG.  31.    The  generator  G  supplies  power  to  the  line  and  the 
motor  M. 

Example  3.  In  Fig.  31,  the  generator  generates  an  electro- 
motive force  of  250  volts.  The  generator  has  a  resistance  of 
L/5  ohms  and  the  line  wires  have  resistances  of  2  ohms  each. 
What  power  is  delivered  to  the  mot/>r  M? 

Power  (to  motor)  =  Current   (through  motor)   x  volt- 

age (across  motor). 

Voltage  across  motor      =  electromotive    force    of    generator 

—  IR  drop  in  generator  and  line 
wires  . 

=  250  -  (fr-7T*»5)  -  (5  X  4)  , 
=  222.5  volts. 

Current  through  motor  =  5  amperes. 
Power  to  motor  =  222.5  X  5 

=  1112.5  watts 
=  1.11  kilowatts. 


ELECTRIC  POWER  AND  ENERGY  55 

It  will  be  noted  that  the  result  in  the  above  has  been 
rounded  off  to  three  figures.  It  would  hardly  be  correct  to 
express  the  result  as  1.1125  kilowatts  for  we  cannot  measure 
as  closely  as  this  by  any  ordinary  means.  The  number  of 
figures  given  in  a  result  should  correspond  to  the  accuracy 
which  can  be  attained  under  the  conditions  of  the  problem. 
Ordinary  voltages,  currents  and  powers  can  be  quickly 
measured  with  the  usual  instruments  with  the  accuracy  of 
about  one-half  of  one  per  cent.  By  using  care  and  special 
instruments  they  can  be  measured  somewhat  more  accu- 
rately. 

25.  Power  Consumed  by  Resistance.  We  have  seen 
that  the  voltage  required  to  force  a  current  of  7  amperes 
through  a  resistance  of  R  ohms  is  IR  volts. 

From  this  relation  we  can  determine  the  power  consumed 
in  sending  a  current  through  a  resistance. 

P  =  IE 
but  E  =  IR, 

Therefore 

P=  PR 

also 


Therefore 

We  thus  have  three  forms  for  the  power  equation 
P  =  IE  =  PR  =  ^- 

The  form  P  =  IE  is  universal,  in  that  P  is  the  power  re- 
ceived or  delivered  when  a  current  of  7  amperes  flows  under 
a  pressure  of  E  volts. 

The  forms  P  =  PR  and  P  =  -p  apply  only  when  #  is 

the  pressure  required  to  send  a  current  of  7  amperes  through 
a  resistance  of  H  ohms. 


56        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Example  4.     How  much  power  is  expended  in  overcoming 
the  resistance  of  the  line  wires  of  Fig.  31? 


Where 

PR  =  power  consumed  by  resistance  R, 

ER  =  voltage  used  in  sending  current  through  R 

=  4  X  5  =  20  volts, 
7     =  current  through  R, 
PR  =  PR  =  5  x  5  X  4  =  100  watts 
or 

_  ^  =  »XM  _  100  watts 

or 

=  I  ER  =  20  X  5  =  100  watts. 

It  may  often  be  necessary  to  apply  both  forms  of  the 
equation  to  one  part  of  a  circuit  in  order  to  determine  how 
the  power  is  consumed  or  expended.  This  happens  when 
power  in  a  given  part  of  a  circuit  is  expended  in  more  than 
one  way. 

Thus  in  the  generator  in  Fig.  30,  the  power  generated  is 
expended  in  two  ways,  —  first,  in  sending  the  5-ampere 
current  through  the  generator  against  the  1.5-ohm  internal 
resistance,  and  second,  in  delivering  current  to  the  outside 
line. 

Example  5.     How  much  power  is  delivered  to  the  line  by 
the  generator  in  Fig.  31? 
Total  power  generated 

P  =  IE 

=  5  X  250 

=  1250  watts. 

Power  consumed  by  internal  resistance 
PR  =  PR 

=  5  X  5  X  1.5 

=  37.5  watts. 
Power  delivered  to  line 

=  P  -  PR  =  1250  -  37.5 

-  1212.5  watts. 


ELECTRIC  POWER  AND  ENERGY  57 

Or  the  power  consumed  by  internal  resistance  may  be  found 
as  follows: 

Voltage    (ER]   used  to  force   current  through   cuiTEStTTF- 
through  internal  resistance  R 

=  IR 
=  5  X  1.5 
=  7.5   volts. 
Power  consumed  in  R 

=  IER 
=  5  X  7.5 
=  37.5  watts. 

In  all  of  the  above  equations  for  computing  the  power 
used  in  the  resistance  R,  note  that  the  voltage  used  is  ER, 
the  voltage  required  to  force  the  current  through  the  re- 
sistance R. 

In  the  case  of  a  motor,  part  of  the  electric  energy  deliv- 
ered to  it  is  used  to  overcome  the  resistance  of  the  windings, 
the  rest  is  transformed  into  mechanical  power,  some  of 
which  is  in  turn  lost  in  friction  and  in  other  losses  to  be 
studied  later. 

Example  6.  If  the  resistance  of  the  motor  in  Fig.  31  is  3 
ohms,  how  much  mechanical  power  is  developed? 

In  Example  3  we  determined  that  1110  watts  were  delivered 
to  the  motor. 

Power  consumed  by  resistance  of  motor 

PR  =  I2XRr 
=5X5X3 

=  75  watts. 

Power  transformed  into  mechanical  power 

P  -   PR  =  1110  -  75 
=  1035  watts 

or  -  =  1.39  horse  power. 

746 

Some  of  this  mechanical  power  would  be  used  in  over- 
coming the  frictional  resistance  to  motion,  etc.,  so  that  the 
motor  would  probably  deliver  about  one  horse  power  which 
could  be  used  in  driving  other  machinery. 


58        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Prob.  1-3.  A  110- volt  motor  having  0.625  ohm  resistance 
receives  4.20  kilowatts  at  its  terminals.  What  mechanical 
horse  power  is  developed  in  the  motor? 

Prob.  2-3.     Fig.  31a  represents  the  electrical  circuit  of  an 
arc  lamp.     The  voltage  across  the  terminals  is  50  volts,  and 
across  the  arc  alone  it  is  40  volts.     The  arc  takes  6.2  amperes. 
The  resistance  of  the  shunt  coil  S  is  380  ohms.     Find 
(a)  The  power  consumed  by  the  lamp, 
(6)  The  resistance  of  the  ballasting  resistance  R, 
(c)    The  power  consumed  by  the  resistances  S  and  R. 


0.09  Ohm 


0.14  Ohm 


0000 


FIG.  31a.  The  cir- 
cuits through  an 
arc  lamp. 


C  D 

FIG.  32.     The  generator  G  supplies  power  to  the 
motor  M  and  the  lamps  L. 


Prob.  3-3.  The  motor  M  in  Fig.  32  has  a  resistance  of  1.03 
ohms  and  takes  20  amperes.  Each  lamp  at  L  has  a  resistance 
of  40  ohms  and  takes  2.42  amperes. 

(a)  What  power  is  taken  by  the  lamp  bank  L? 

(6)  What  power  is  taken  by  the  motor? 

(c)  How   much   electrical   power   is    converted   into    me- 

chanical power  in  the  motor? 

(d)  How  much  power  is  lost  in  the  line  wires? 

26.  Measurement  of  Electric  Power.  Since  in  direct- 
current  measurements  the  power  is  the  product  of  the 
volts  and  amperes,  the  combination  of  a  voltmeter  and  an 
ammeter  may  be  used  to  measure  the  power  in  any  part  of 


ELECTRIC  POWER  AND  ENERGY 


59 


a  circuit.  However,  these  two  instruments  have  been  com- 
bined in  one  instrument  called  a  wattmeter. 

The  current  is  led  through  the  field  coils  FF  (Fig.  33) 
making  a  magnetic  field  proportional  to  the  strength  of  the 
current.  The  voltage  is  impressed  on  the  coil  P  which  has 
a  high  resistance  R  in  series 
with  it.  The  turning  tendency 
of  the  coil  P  is,  therefore,  pro- 
portional to  the  product  of  the 
current  by  the  voltage,  that  is, 
to  the  power  in  a  direct-current 
circuit. 

An   instrument   of   this   type  T 

„      ,         -  .     ,  FIG.  33.    The  wiring  diagram 

usually  has  four  terminals,  two  of  a  wattmeter 

for  the  current   leads   and   two 

for  the  voltage  leads.  The  greatest  care  must  always  be 
exercised  not  to  confuse  these  two  sets  of  terminals,  for 
inasmuch  as  the  current  coils  are  of  very  low  resistance, 
the  instrument  is  ruined  if  the  mistake  is  made  of  connect- 
ing the  voltage  leads  to  these  coils. 

Fig.  34  shows  the  proper  method  of  connecting  a  watt- 
meter to  indicate  the  power  taken  by  the  bank  of  lamps  (B). 


FIG.  34. 


The  wattmeter  measures  the  power  taken  by  the  lamp 
bank  B. 


The  reason  that  a  separate  instrument  is  used  for  a  watt- 
meter, instead  of  simply  a  voltmeter  and  an  ammeter,  is  as 
follows.  In  a  circuit  in  which  the  current  is  steady,  that  is, 
in  a  direct-current  circuit,  the  power  is  equal  to  the  product 
of  E  and  7.  When  the  current  and  voltage  in  the  circuit 
are  varying  this  is  not  true  unless  they  are  varying  in  ex- 


60        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

actly  similar  manner.  The  relation  is,  however,  true  for 
the  instantaneous  voltage,  current  and  power  in  any  circuit. 
When  used  in  a  circuit  in  which  the  current  and  voltage  are 
varying,  the  wattmeter  gives  a  deflection  corresponding  to 
the  average  of  these  instantaneous  values,  and  hence  shows 
the  actual  average  power  in  the  part  of  the  circuit  to  which 
it  is  connected.  This  cannot  be  done  except  where  the 
wattmeter  arrangement  of  coils  is  used.  Hence  while  a 
wattmeter  is  not  necessary  in  order  to  measure  direct-cur- 
rent power,  it  is  the  only  convenient  way  in  which  alter- 
nating-current power  can  be  measured. 

27.  Electric  Energy.  Since  power  is  the  rate  of  doing 
work,  the  total  amount  of  work  done  or  energy  expended 
in  a  given  time  is  the  product  of  the  power  and  the  time. 
Thus  a  25-horse-power  steam  engine  running  at  full  load  for 
4  hours  does  25  X  4  or  100  horse -power-hours  of  work. 
Similarly  a  25-kilowatt  generator  running  at  full  load  for 
4  hours  delivers  25  X  4  or  100  kilowatt-hours  of  electric 
energy.  A  kilowatt-hour  is  the  energy  delivered  or  re- 
ceived by  an  appliance  in  one  hour  if  the  power  is  main- 
tained at  one  kilowatt. 

Similar  units  of  electric  energy  are  the  watt-hour  and  the 
watt-second  or  joule.  The  values  of  these  are  evident  from 
the  names. 

1  kilowatt-hour  =  1.34  horse-power-hours. 
1  watt-second  (joule)  =  0.737  foot-pound. 

Prob.  4-3.  If  it  costs  a  total  of  $0.0126  per  kilowatt-hour 
to  generate  electric  energy,  at  what  price  must  it  be  sold  per 
horse-power-year  in  order  to  realize  a  10  %  profit?  Assume  the 
power  is  to  be  used  24  hours  per  day  and  365  days  per  year  in 
a  chemical  plant. 

Prob.  5-3.  A  generator  supplies  for  8  hours  per  day  40  kilo- 
watts at  110  volts  to  a  motor  load  situated  500  feet  from  the  gen- 
erator. The  resistance  of  the  line  wires  is  0.049  ohm  per  1000 
feet.  It  costs  $0.056  per  kilowatt-hour  to  generate  electric  en- 
ergy. How  much  money  would  be  saved  in  a  year  of  300  days  by 
doubling  the  voltage  of  the  motors  and  generator  and  supplying 
the  motors  with  the  same  power? 


ELECTRIC  POWER  AND  ENERGY 


61 


Prob.  6-3.  How  many  25-watt  lamps  can  be  used  (on  the 
average)  for  2j  hours  per  night  if  the  monthly  bill  is  not  to 
exceed  $3.20?  The  price  of  electric  energy  is  10  cents  per 
kilowatt-hour;  a  month  is  to  be  reckoned  as  having  30  days. 

28.  Heat  Energy  of  Electricity.  We  have  seen  that  when 
a  current  /  is  made  to  flow  through  a  resistance  R,  energy 
is  expended  at  the  rate  of 
PR  watts.  This  energy 
is  all  transformed  into 
heat  just  as  all  mech- 
anical energy  which  a 
machine  uses  in  overcom- 
ing friction  is  transformed 
into  heat.  This  fact  is 
strikingly  called  to  mind 
in  Fig.  35,  which  is  a 
photograph  of  a  large 
transformer.  Note  that 
a  special  arrangement  of 
pipes  is  necessary  so  that 
the  oil  in  the  case  can 
circulate  and  be  cooled 
by  the  air  passing  over 
the  pipes.  Much  of  the 
heat  thus  dissipated  origi- 
nates in  the  coils  and  is 
due  to  the  passage  of  the 
electric  current  through 
the  resistance  of  the  coils. 

Thus  the  energy,  re- 
presented by  the  expres- 
sion (PRi),  always  appears  as  heat  and  can  be  expressed 
equally  well  in  heat  units  as  calories  or  in  electric  units  as 
watt-seconds.  It  is  necessary  to  know  merely  the  relation 
of  a  calorie  to  a  watt-second.  Careful  measurements  have 
determined  that  one  watt-second  equals  0.24  calorie. 


FIG.  35.  A  typical  self-cooled  radiator- 
type  transformer  of  over  2000  kw. 
capacity.  Westinghouse  Elec.  &  Mfg. 
Co. 


62         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Thus 

H  =  0.24  PRt, 
where 

H  =  heat  generated  in  calories, 
I  =  current  in  amperes, 
R  =  resistance  in  ohms, 
t  =  tune  in  seconds. 


Of  course  the  expression  IERt  or  —5-    can  be  used  in  the 

rt 

place  of  I2Rt,  providing  ER  represents  just  the  voltage  re- 
quired to  force  the  current  /  through  the  resistance  R. 

Example  7.  An  electric  water  heater  designed  to  operate 
on  110- volt  lines  has  a  resistance  of  15  ohms.  How  long  will 
it  take  to  raise  the  temperature  of  a  cup  of  water  containing 
250  grams  from  10°  C.  to  90°  C.,  assuming  no  loss  of  heat 
from  the  cup,  and  neglecting  the  specific  heat  of  the  cup  itself? 

Heat  required  to  raise  250  grams  of  water  80°  C. 

H  =  80  X  250 

=  20,000  calories; 
also 

t  :•    •;  *  =  °-24?- 

Therefore 

20,000  =  110  X  "°  ><  °-24' 
15 

20,000  X  15 
"  110  X  110  X  0.24 
=  103.3  seconds 
=  1.72  minutes. 

29.  Efficiency  of  Transmission.  Regulation.  Since  some 
energy  is  always  transformed  into  heat  whenever  an  elec- 
tric current  flows,  no  scheme  of  transmission  can  have  an 
efficiency  of  100  %.  In  other  words,  the  PR  loss  is  always 
present  in  the  conductors  of  the  transmission  line  whenever 
any  electric  energy  is  being  transmitted. 

The  efficiency  of  transmission  may  be  defined  as  the 


ELECTRIC  POWER  AND  ENERGY  63 

ratio  of  the  useful  output  to  total  input  of  the  line.  In  the 
low  voltage  systems  (under  5000  volts)  generally  used  for 
short-distance  transmission  of  direct  currents,  the  PR  loss 
is  practically  the  only  loss  in  the  line.  Computing  the 
efficiency  of  transmission  of  such  a  line  is  therefore  easily 
done. 

Example  8.     In  the  system  shown  in  Fig.  36,  the  appliances 
at  B  draw  16  amperes  and  those  at  A,  20  amperes.     The  re- 

0.125  Ohms  0.1875  Ohms 


Q.125  Ohma  I  0.1875  Ohms 


FIG.  36.    A  transmission  system  supplying  power  to  two  groups  of 
lamps  from  one  generator. 

sistance  of  line  and  return  between  the  generator  G  and  group 
A  is  0.25  ohm;  between  A  and  B  0.375  ohm.  What  is  the 
efficiency  of  transmission? 

Total  input  into  line  =  125  X  (20  +  16) 

=  4500  watts. 

PR  loss  between  G  and  A    =  362  X  0.25 

=  324  watts. 
PR  loss  between  A  and  B  =  162  X  0.375 

=  96  watts. 

Total  line  loss  =  96  +  324 

=  420  watts. 

Useful  output  =  4500  -  420 

=  4080  watts. 

,.,„.  .  output      4080       _n  .07 

Efficiency  =    .    K      =  -  =  90.7%. 
input        4500 

As  a  check  we  may  compute  the  output  more  directly  as 
follows: 

Line  drop  G  to  A  =  IR 

=  36  X  0.25     — 
=  9  volts. 

Voltage  across  A  =125-9 
=  116  volts. 


64        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Power  used  by  A  =  116  X  20 

=  2320  watts. 
Line  drop  A  to  B  =  IR 

=  16  X  0.375 

=  6  volts. 
Voltage  across  B  =  116  -  6 

=  110  volts. 
Power  used  at  B  =  110  X  16 

=  1760  watts. 
Total  power  used  by  A  and  B  =  2320  +  1760 

=  4080  watts, 
(check) 

The  regulation  of  a  transmission  line  is  a  measure  of  the 
drop  in  voltage  when  the  line  is  loaded.  It  is  defined  as 
the  rise  in  terminal  voltage  when  the  load  is  changed  from 
full  load  to  no  load,  divided  by  the  full-load  voltage.  The 
regulation  of  a  generator  or  an  alternating-current  trans- 
former is  defined  in  exactly  similar  manner.  The  regulation 
of  a  transmission  line  should  be  low  in  order  not  to  subject 
the  load  to  undesirable  fluctuations  of  voltage. 

Example  8  (continued).  Assuming  the  load  above  to  be 
full  load  for  the  transmission  line,  we  may  compute  the  reg- 
ulation as  follows: 

Voltage  at  B,  full  load  =  110  volts. 
Voltage  at  B,  no  load 

(same  as  generator  voltage)  =  125  volts. 
Change  of  voltage,  no  load  to  full  load,  =125  —  110 

=  15  volts. 

Regulation  =  -^j-  =  0.136  =  13.6%. 

In  making  this  computation  we  neglect  any  possible 
change  of  generator  voltage  with  load,  as  we  wish  to  com- 
pute the  regulation  of  the  transmission  line  only. 

Prob.  7-3.  An  electric  circuit  has  the  form  of  a  loop,  i.e., 
each  conductor  is  connected  to  the  power  station  busbar  of 
like  sign  at  both  ends.  The  loop  is  1.25  miles  long  and  each 
conductor  has  a  resistance  of  0.26  ohm  per  mile.  The  two  bus- 
bars at  the  power  station  have  a  potential  difference  of  575 
volts.  The  loop  circuit  supplies  two  shop  buildings,  one  at  a 


ELECTRIC  POWER  AND  ENERGY 


65 


distance  of  1800  feet  from  one  end  of  the  loop  taking  400  am- 
peres and  the  other  at  a  distance  of  2600  feet  from  the  other 
end  of  the  loop  taking  560  amperes.  What  is  the  voltage 
between  lines  at  each  shop? 

(a)  What  is  the  efficiency  of  transmission? 

(6)  Determine  the  voltage  at  each  shop  and  the  efficiency  of 
transmission  if  that  portion  of  the  loop  between  the  two  shops 
were  omitted. 

Prob.  8-3.  The  field  coils  of  a  certain  generator  contain 
20  pounds  of  copper  (specific  heat  0.095).  The  weight  of  the 
insulation  is  negligible.  The  resistance  of  the  field  coils  is 
250  ohms.  How  fast  would  the  temperature  rise  if  there  were 
no  cooling  by  convection  or  radiation  and  a  current  of  2.4 
amperes  were  flowing  in  the  coils? 

Prob.  9-3.  In  Fig.  37,  station  I  takes  50.0  amperes  and 
station  II  takes  65.0  amperes.  The  resistance  of  AB  and  of 
EC  each  equals  0.0412  ohm.  The  resistance  of  CD  and  of 


ii 


M  K  F  E 

FIG.  37.    A  transmission  system  where  two  generators  are  used  to 
supply  power  to  two  stations. 


MK  each  equals  0.0330  ohm.  The  resistance  of  EF  and  of  KF 
each  equals  0.0214  ohm.  The  terminal  voltage  of  Gi  is  121 
volts  and  of  G2  is  125  volts. 

(a)  What  is  the  voltage  regulation  at  station  I? 

(6)  What  is  the  voltage  regulation  at  station  II? 

(c)   What  is  the  efficiency  of  transmission? 

Prob.  10-3.  If  generator  G2  were  removed  from  the  line 
in  Fig.  37  and  all  other  data  remained  as  in  Prob.  9-2,  compute 

(a)  The  voltage  regulation  of  station  I, 

(6)  The  voltage  regulation  of  station  II, 

(c)   The  efficiency  of  transmission. 


66        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

30.  The  Three-wire  System  of  Transmission.  In  order 
to  transmit  a  given  amount  of  power  over  a  line  with  as 
small  PR  loss  as  practicable,  it  is  necessary  to  make  the 
current  as  small  as  possible.  This  can  be  done  by  increasing 
the  voltage  of  transmission.  Thus  10,000  watts  may  be 
transmitted  in  the  form  of  100  amperes  at  100  volts  pressure 
or  of  50  amperes  at  200  volts  pressure,  or  in  any  other  form 
the  factors  of  which  multiplied  together  equal  10,000  watts. 
But  if  transmitted  as  50  amperes  at  200  volts,  the  line  loss 

502 
would  be  only  -^-2  or  J  as  much  as  if  it  were  transmitted  as 


100  amperes  at  100  volts,  the  same  line  being  used  in  both 
cases.  Thus  doubling  the  voltage  has  decreased  the  line 
loss  to  one-quarter  as  much. 

For  a  given  amount  of  power  transmitted  over  a  given 
line: 

The  line  loss  is  proportional  to  the  square  of  the  line 

current. 
The  line  current  is  inversely  proportional  to  the  line 

voltage. 

Therefore  the  line  loss  is  inversely  proportional  to  the 
square  of  the  voltage  of  transmission. 

Prob.  11-3.  At  a  certain  place  there  are  available  30  kw. 
of  electric  power.  The  line  wires  from  this  place  to  the  load 
have  a  resistance  of  1.5  ohms  each.  Plot  a  curve  between 
efficiency  of  transmission  and  voltage  at  sending  end  for  values 
of  voltage  varying  from  100  volts  to  10,000  volts. 

In  direct-current  systems,  only  a  limited  advantage  can 
be  taken  of  the  higher  efficiency  of  high-  voltage  transmission. 
While  motors  can  be  constructed  for  operation  at  fairly 
high  voltages,  nearly  all  the  other  ordinary  electrical  ap- 
pliances are  operated  most  easily  at  about  110  volts. 

This  is  particularly  true  of  electric  incandescent  lamps. 
A  lamp  constructed  to  operate  at  220  volts  or  550  volts,  for 
instance,  would  be  expensive  and  of  short  life. 

With  alternating  currents  we  are  not  subject  to  limitation 


ELECTRIC  POWER  AND  ENERGY 


67 


in  this  regard.  Alternating-current  power  can  be  trans- 
formed from  one  voltage  to  another  at  will,  by  reliable 
stationary  apparatus,  and  with  little  loss.  The  alternating- 
current  transformer  henc§  makes  possible  the  transmission 
of  large  amounts  of  power  over  distances  of  even  200  miles 
and  at  voltages  as  high  as  220,000  volts.  At  the  point 
wher&  it  is  utilized  it  may  be  transformed  down  to  a  safe 
convenient  voltage  around  110  volts.  In  fact,  it  is  usually 
transformed  several  times  between  generator  and  user. 
The  distance  to  which  power  can  be  transmitted  econom- 
ically depends  upon  the  amount  of  power  to  be  carried  and 
the  voltage  used.  Direct-current  transmission  is  hence 
limited  to  comparatively  short  distances,  although  many 
miles  of  track  are  covered  by  a  single  station  in  direct- 
current  railway  practice  where  voltages  of  3600  and  even 
higher  are  used. 


N 


FIG.   38.     Two  generators  in  series  used  to  supply  a  three-wire  system 
to  which  are  attached  110-volt  lamps  and  a  220-volt  motor. 

Since  there  is  no  direct-current  transformer,  direct-cur- 
rent power  must  be  transmitted  at  practically  the  same 
voltage  at  which  it  is  used.  Much  advantage,  however,  is 


68        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

gained  by  using  three  wires  instead  of  two.  Thus  in  Fig. 
38,  two  110- volt  generators  are  shown  connected  in  series. 
Therefore,  across  the  outside  conductors  A  and  B  there  are 
impressed  220  volts,  while  only  110  volts  exist  between  either 
of  these  conductors  and  the  neutral  wire  N.  Accordingly 
the  lamps  have  practically  110  volts  across  them,  while 
the  motor  has  about  220  volts.  This  system  therefore  com- 
bines the  advantage  of  affording  both  110  volts  and  220 
volts,  and  yet  having  nearly  the  efficiency  of  a  220-volt  line. 

The  greatest  efficiency  is  secured  when  the  loads  on  both 
sides  of  the  neutral  are  "  balanced  "  so  that  no  current 
flows  in  the  neutral.  This  is  the  condition  which  should 
be  approximated. 

Instead  of  connecting  two  generators  in  series,  it  is  cus- 
tomary to  use  a  special  three-wire  generator,  or  a  large  gen- 
erator and  a  balancer  set.  These  machines  are  fully  de- 
scribed in  texts  on  electrical  machinery. 

Prob.  12-3.  Assuming  that  the  cross-section  of  a  con- 
ductor must  be  directly  proportional  to  the  current  it  is  to 
carry,  compare  the  weight  of  copper  required  by  a  three-wire 
system  with  the  weight  required  by  a  two-wire  system  trans- 
mitting the  same  amount  of  power.  All  three  wires  in  the  three- 
wire  system  are  to  be  the  same  size. 

Prob.  13-3.     Each  lamp  in  Fig.  39  takes  1.2  amperes.     Find: 

(1)  Line  drop, 

(2)  Voltage  across  each  set  of  lamps, 

(3)  Efficiency  of  transmission. 

Prob.  14-3.  If  the  lamp  bank  marked  S,  Fig.  39,  is  turned 
off,  and  the  resistance  of  the  other  lamps  remains  the  same 
as  in  Prob.  13-3,  find: 

(1)  Line  drop, 

(2)  Voltage  across  remaining  sets  of  lamps, 

(3)  Efficiency  of  transmission. 

Prob.  16-3.  Assume  that  each  lamp  in  Fig.  40  takes  2 
amperes  and  that  the  resistance  of  the  lamps  remains  constant. 
The  brush  potential  of  the  generator  is  220  volts.  Each  small 
machine  maintains  a  terminal  voltage  of  110  volts.  See  Prob. 
20-3.  Find: 


ELECTRIC  POWER  AND  ENERGY 


69 


(a)  Line  drop  in  each  section, 

(6)  Voltage  across  each  set  of  lamps, 

(c)   Efficiency  of  transmission. 

Prob.  16-3.  If  a  break  occurs  in  the  neutral  between  0  and 
S,  Fig.  40,  what  would  be  the  values  of  (a),  (6)  and  (c),  Prob. 
15-3?  Lamp  resistance  remains  as  in  Prob.  15-3. 

MA  B 


0.2  Ohm 


0.2  Ohm 


0.3  Ohm 


0.3  Ohm 


2V  0.2Ohm  £  O.SOhm 

FIG.  39.    Two  generators  feeding  a  balanced  three-wire  system. 


FIG.  40.    A  generator  with  a  balancer  set  supplying  power  to  a 
three-wire  system. 

Prob.  17-3.  If  a  break  occurs  in  the  neutral  between  S 
and  V,  Fig.  40,  what  would  be  the  values  of  (a),  (b)  and  (c), 
Prob.  15-3? 


SUMMARY    OF   CHAPTER  III 

Direct-current  electric  power  is  measured  in  watts  and  can 
be  computed  from  the  equations 
P  =  IE 
=  I2R 

=  ?! 

~R* 

E2 
The  quantities  I2R  and  —    represent  power  consumed  by  the 

R 

resistance  R,  which  appears  as  heat  in  the  resistor. 

The  units  of  electric  energy  are  the  watt-second,  commonly 
called  the  joule,  and  the  kilowatt-hour. 

The  watt-second  equals  0.24  calory.  Thus  the  heat  lib- 
erated by  an  electric  current  can  be  obtained  from  the  equation 

H  =  0.24  l2Rt, 

where 

H  is  in  calories, 
I  is  in  amperes, 
R  is  in  ohms, 
t  is  in  seconds. 

The  efficiency  of  electric  transmission  is  the  fraction 

power  delivered  to  load 
power  received  from  generator 

In  the  transmission  of  a  given  amount  of  power,  the  line  loss 
is  inversely  proportional  to  the  square  of  the  voltage  of  trans- 
mission. Thus  the  higher  the  voltage,  the  higher  the  efficiency 
as  far  as  it  depends  upon  the  I2R  loss  of  the  line. 

Because  of  the  higher  voltage  of  transmission  available, 
the  three-wire  system  is  in  general  use  for  direct-current 
distribution. 

The  voltage  regulation  of  a  line  is 

no-load  voltage  —  full-load  voltage 
full-load  voltage 

These  voltages  are  measured  at  the  point  where  the  load  is 
attached  to  the  line. 

70 


PROBLEMS   ON   CHAPTER  HI 

Prob.  18-3.  A  shunt  dynamo  acts  as  a  generator  and  sup- 
plies 4710  watts  of  power  to  a  lighting  system  connected  across 
its  terminals.  In  addition  the  armature  supplies  the  exciting 
current  to  the  field  coils  which  have  a  resistance  of  136.4  ohms. 
The  armature  resistance  is  0.12  ohm  and  the  terminal  voltage 
124.  (a)  What  is  the  armature  current?  (b)  What  e.m.f. 
is  generated  in  the  armature?  (c)  What  power  is  lost  as  heat 
in  the  windings? 

Prob.  19-3.  A  shunt  generator  supplies  power  to  charge  a 
storage  battery.  The  armature  of  the  generator  has  a  resist- 
ance of  0.32  ohm.  The  shunt  field  winding  of  the  generator 
has  a  resistance  of  140  ohms.  Each  of  the  two  conductors 
leading  to  the  battery  has  a  resistance  of  0.2  ohm.  The  bat- 
tery exerts  by  a  chemical  action  an  e.m.f.  of  112  volts  and  has 
an  internal  resistance  of  0.08  ohm.  What  e.m.f.  must  be  set 
up  in  the  armature  of  the  generator  to  charge  the  battery  at 
the  rate  of  24  amperes?  If  the  rotational  losses  of  the  generator 
at  this  state  are  160  watts,  what  mechanical  power  must  be 
employed  to  drive  the  generator?  What  percentage  of  this 
mechanical  energy  is  stored  chemically  in  the  battery? 

Prob.  20-3.  Fig.  41  represents  the  method  of  supplying  a 
three-wire  system  from  a  two-wire  source  by  the  aid  of  a  bal- 
ancer set.  At  the  original  two-wire  source  a  constant  potential 
difference  of  236  volts  is  maintained.  The  balancer  set  com- 
prises twin  dynamo  machines  whose  armatures  A  and  B  are 
mechanically  coupled  and  electrically  in  series  between  the 
outside  wires.  The  middle  wire  is  led  out  from  the  intermediate 
point.  Each  armature  of  the  balancer  set  has  a  resistance  of 
0.028  ohm.  The  line  resistances  are  as  indicated.  When  the 
three-wire  system  is  unloaded,  the  balancer  set  runs  idle  as  two 
motors  in  series.  When  one  side  of  the  three-wire  system  is 
loaded  more  heavily  than  the  other,  the  machine  on  the  heavily 
loaded  side  operates  as  a  generator  and  is  driven  by  the  other 
machine  running  as  a  motor.  The  field  windings  of  the  bal- 
ancer set,  not  shown  in  the  sketch,  are  to  be  so  regulated  that 
equal  voltages  are  to  exist  between  R  and  S  and  between  S 
and  T.  The  load  on  one  side  is  80  amperes  and  the  load 

71 


72        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

on  the  other  10  amperes.  The  rotational  losses  of  the  balancer 
set  amount  to  472  watts  and  are  supplied  from  the  primary 
two-wire  source. 


4MAM/WWV 

0.018  Ohm 


•A/WMMW 

0.03  Ohm 


0.018  Ohm 

-AAMMAMA/V 


— 'VVVVVVVVVVV -»*' 

FIG.  41.     A  diagrammatic  representation  of  a  generator  with  a 
balancer  set  supplying  power  to  a  three- wire  system. 


(a)  What  e.m.f.'s  must  be  induced  in  the  two  armatures 
under  the  circumstances  above  described? 

(6)  What  are  the  voltages  between  R  and  S,  and  S  and  T7? 

(c)  What  are  the  voltages  between  R  and  S,  and  S  and  T 
when  the  load  on  each  side  is  80  amperes? 

Prob.  21-3.  If  in  Prob.  20-3  the  load  is  removed  between 
the  lines  R  and  S,  what  will  be  the  voltages  generated  in  the 
two  armatures  and  what  will  be  the  voltage  across  line  ST 
at  a  load  of  80  amperes? 

Prob.  22-3.  A  garage  storage-battery  set  has  been  proposed 
having  the  following  points :  1st,  a  d-c.  generator  having  a 
70- volt  and  a  105- volt  adjustment,  and  an  internal  resistance 
under  these  two  conditions  of  0.15  ohm;  2nd,  a  smaller  ma- 
chine having  two  armature  coils  capable  of  generating  5  volts 
each.  The  internal  resistance  of  each  5-volt  coil  is  0.04  ohm. 
These  may  be  arranged  so  that  they  separately  buck  or  boost 
the  main  generator  voltage.  In  this  way  combinations  can 
be  obtained  to  give  voltage  from  60  to  115  volts  in  5-volt  steps 
with  the  exception  of  85  volts  and  90  volts.  The  rotational 
losses  in  the  two  machines  total  400  watts.  It  is  desired  to 
charge  7  batteries  of  average  back  e.m.f.  of  11.2  volts  at  a  40- 
ampere  rate  for  8  hours.  Determine  the  arrangement  of  the 


ELECTRIC  POWER  AND  ENERGY  73 

coils  of  this  system  so  that  the  charging  rate  nearest  to  40  am- 
peres can  be  obtained.  Under  this  condition  find  the  saving 
over  a  system  using  a  plain  11 5- volt  generator  with  resistance  in 
series  with  the  batteries  to  control  the  current.  Rotational 
loss,  300  watts.  The  cost  of  electric  energy  may  be  taken  as 
three  cents  per  kilowatt-hour.  The  average  battery  internal 
resistances  is  0.03  ohm.  (Electrical  Review,  VoLGO.) 

Prob.  23-3.  In  Prob.  22-3  what  would  be  the  overall  charg- 
ing efficiency  for  both  systems  and  what  would  be  the  losses? 

Prob.  24-3.  A  certain  railroad  electrification  is  five  miles 
long.  Power  is  sent  out  over  a  1,000,000-circular-mil  feeder 
whose  resistance  is  0.057  ohm  per  mile.  The  feeder  is  tied  at 
both  ends  and  every  half  mile  to  a  #  0000  trolley  wire  whose 
resistance  is  0.26  ohm  per  mile.  The  resistance  of  the  return 
rail  circuit  is  0.02  ohm  per  mile.  The  generator,  which  is 
separately  excited,  has  an  armature  resistance  of  0.05  ohm  and 
an  induced  electromotive  force  of  600  volts.  At  the  distant 
end  of  the  line  is  a  storage  battery  whose  resistance  is  0.3  ohm. 
A  train  taking  700  amperes  is  three  miles  from  the  generator 
end  of  the  line.  If  the  generator  is  supplying  600  amperes,  find : 
(a)  Voltage  at  generator  end  of  the  line, 
(6)  Voltage  at  car, 

(c)  Voltage  at  battery  end  of  line, 

(d)  Electromotive  force  of  battery, 

(e)  Number  of  coulombs  delivered  by  the  battery  in  ten 
seconds, 

(/)  Kilowatts  input  to  car. 

Prob.  26-3.  If  the  controller  on  the  train  (Prob.  24-3) 
should  be  shut  off  so  that  no  current  was  taken  by  the  train, 
how  much  current  (if  any)  would  flow  into  the  battery? 

Prob.  26-3.  Two  railroad  substations  are  6  miles  apart. 
The  trolley  wire  is  reinforced  by  a  heavy  copper  feeder  and  the 
two  in  parallel  are  equivalent  to  a  conductor  whose  resistance 
is  0.018  ohm  per  1000  feet.  The  two  rails  serve  in  parallel  as 
a  return  conductor  and  have  a  combined  resistance  of  0.006 
ohm  per  1000  feet.  At  a  distance  of  2.5  miles  from  one  end  of 
the  line  a  train  is  running  at  a  speed  of  30  miles  per  hour  against 
an  opposing  force  of  7200  pounds.  The  electrical  equipment  of 
this  train  has  an  efficiency  of  86  %.  The  voltage  between  bus- 
bars at  the  substations  is  700.  What  is  the  voltage  between 
trolley  and  track  at  the  train?  What  current  does  the  train 
receive  from  each  substation? 


74        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  27-3.  If  a  train  moves  from  one  end  of  the  railroad 
referred  to  in  Prob.  26-3  to  the  other  end  and  requires  at  all 
times  a  current  of  800  amperes,  by  what  equation  may  the 
voltage  at  the  train  be  expressed  in  terms  of  its  distance,  X, 
from  the  end  of  the  line?  Draw  a  curve  to  indicate  approxi- 
mately the  general  nature  of  the  relation  between  the  position 
of  the  train  and  the  voltage  acting  on  it. 

Prob.  28-3.  A  railroad  line  is  supplied  with  power  from 
both  ends.  At  one  end  a  power  station  maintains  a  constant 
voltage  between  positive  and  negative  busbars  of  600.  At  the 
other  end  is  a  storage  battery  which  sets  up  an  e.m.f.  of  540 
volts  by  chemical  action.  The  internal  resistance  of  this 
battery  is  0.08  ohm.  The  positive  conductor  is  a  trolley  wire 
in  parallel  with  a  feeder,  the  two  having  a  combined  resistance 
of  0.098  ohm  per  mile.  The  negative  conductor  consists  of  the 
two  track  rails  in  parallel  and  has  a  resistance  of  0.032  ohm  per 
mile.  The  total  length  of  the  line  is  7.2  miles.  If  only  one 
train  is  running  and  is  located  4  miles  from  the  power  station, 
what  voltage  is  applied  to  it  when  it  takes  200  kilowatts? 

Prob.  29-3.  A  generator  supplies  power  to  a  trolley  load. 
Three  miles  from  the  generating  station  is  a  storage  battery 
connected  between  trolley  and  rail.  Two  miles  beyond  the 
storage  battery  is  a  car  taking  50  kilowatts.  The  trolley 
wire  is  #  0000  having  a  resistance  of  0.26  ohm  per  mile.  The 
resistance  of  the  rail  return  (including  both  rails)  is  0.04  ohm 
per  mile.  The  e.m.f.  and  internal  resistance  of  the  storage 
battery  are  respectively  580  volts  and  0.5  ohm.  If  the  battery 
is  discharging  at  the  rate  of  40  amperes,  calculate  the  current 
taken  by  the  car,  the  voltage  at  the  car,  the  current  delivered 
by  the  generator  and  the  voltage  at  the  generating  station. 

Prob.  30-3.  How  far  from  the  generating  station  in  Prob. 
29-3  would  the  car  be  if  the  storage  battery  were  just  floating 
on  the  line  (neither  charging  nor  discharging)?  The  car  still 
takes  50  kilowatts. 

Prob.  31-3.  How  many  kilowatts  must  be  supplied  to  an 
electric  steel  furnace  which  is  to  deliver  1  ton  of  steel  per  hour? 
Consider  10  %  of  the  heat  to  be  lost  in  radiation. 

Average  specific  heat  of  steel  =  0.167 
Average  temperature  of  fusing  point  =  2400°  F. 
Average  latent  heat  =  50  B.t.u.  per  Ib. 


ELECTRIC  POWER  AND  ENERGY 


75 


Prob.  32-3.  At  5  cents  per  kilowatt-hour,  what  would  be 
the  expense  per  day  of  10  hours  for  heating  a  room  containing 
15  persons?  Allow  30  cubic  feet  of  air  per  minute  per  person. 
Air  enters  at  40°  F.  and  is  heated  electrically  to  70°  F.  Spe- 
cific heat  of  air  (constant  pressure)  is  0.237.  Average  weight 
of  air  is  0.08  pound  per  cubic  foot.  Allow  15  %  loss  in  radiation, 
etc. 

Prob.  33-3.  A  water  rheostat  is  used  as  a  load  during  a  test 
on  a  25-kilowatt  generator.  The  water  flows  into  the  tank  of 
the  rheostat  at  a  temperature  of  54°  F.  and  flows  out  at 
170°  F.  How  much  water  per  hour  is  used  if  the  generator 
is  kept  at  full  load?  Neglect  radiation  and  evaporation. 


0.26  Ohm 


0.39  Ohm 


0.1 3  Ohm 


H  0.04  Ohm        K      0.06  Ohm      F      0.02  Ohm    E 

FIG.  41a.    Trolley  cars  fed  from  both  ends  of  the  line. 

Prob.  34-3  Generator  Gz  in  Fig.  4la  maintains  a  terminal  volt- 
age of  555  volts.  What  is  the  efficiency  of  transmission  when  car 
No.  I  takes  300  amperes  and  car  No.  II,  200  amperes?  The 
resistances  of  the  different  sections  of  the  wire  and  track  are  as 
marked. 


CHAPTER  IV 
THE   COMPUTATION   OF   RESISTANCE 

The  resistance  of  a  conductor  is  directly  proportional  to 
its  length  and  inversely  proportional  to  its  cross-sectional 
area.  If  therefore  the  resistance  of  a  conductor  of  given 
material  of  unit  length  and  unit  cross-section  is  known,  it  is 
possible  to  find  the  resistance  of  a  conductor  of  the  same 
material  whatever  its  length  or  cross-section. 

31.  Resistivity.  Expressed  as  an  equation  the  above 
relation  is 


where 

R  =  the  resistance  of  the  conductor  in  ohms, 
I  =  the  length  of  the  conductor, 
A  =  the  cross-section  area  of  the  conductor, 
p  —  the  resistance  of  a  conductor  of  unit  length  and 
unit  cross-section,  called  the  resistivity  of  the 
material. 

The  unit  length  chosen  for  international  use  is  one  centi- 
meter and  the  unit  area  of  a  cross-section  is  one  square 
centimeter.  The  resistance  of  such  a  centimeter  cube  is 
called  the  resistivity  in  ohm-centimeters.  Of  ordinary  an- 
nealed-copper  wire,  at  20°  C.,  determined  from  careful  tests 
upon  samples  of  the  present-day  grades  in  common  use,  the 
resistivity  is  0.00000172  ohm-centimeter,  usually  stated  as 
1.72  microhm-centimeters. 


76 


THE  COMPUTATION  OF  RESISTANCE  77 

Example  1.  What  is  the  resistance  of  an  annealed-copper 
strap  45  centimeters  long  with  a  cross-section  of  3  X  0.5  cen- 
timeters? 


1.72  X  45  . 

=  —  —  -  —  =51.o  microhms. 
U.o  X  o 

Prob.  1-4.  Prove  by  Ohm's  and  KirchhofFs  Laws  that  the 
resistance  of  a  conductor  is  directly  proportional  to  the  length 
and  inversely  proportional  to  the  cross-section  area. 

32.  Resistance  per  Mil-Foot.  Since,  in  America,  the 
length  of  a  conductor  is  usually  measured  in  feet,  it  is  cus- 
tomary to  express  /  in  equation  (1)  in  feet.  Similarly,  since 
conductors  are  usually  made  circular  in  cross-section,  it  is 
customary  to  use  circular  rather  than  square  measure  in 
denoting  the  area  of  the  cross-section.  The  unit  of  area 
chosen  is  the  circular  mil,  which  is  the  area  of  a  circle  one 
mil  (one  thousandth  inch)  in  diameter.  The  area  of  a 
circle  varies  as  the  square  of  the  diameter.  Since  the  area 
of  a  circle  one  mil  in  diameter  is  one  circular  mil,  the  area  of 
any  circle  in  circular  mils  equals  the  square  of  its  diameter 
in  mils.  A  wire  having  a  length  of  one  foot  and  a  cross-sec- 
tion area  of  one  circular  mil  is  called  a  mil-foot  wire. 

It  is  the  resistance  of  this  mil-foot  wire  expressed  in  ohms 
which  is  used  for  the  value  of  p  in  the  above  equation  when 
the  length  of  the  conductor  is  expressed  in  feet  and  the 
cross-section  area  in  circular  mils. 

For  this  reason  the  equation  is  often  written 


where 

p  =  the  resistance  per  mil-foot, 
I  =  the  length  in  feet, 
d  =  the  diameter  in  mils. 

The  resistance  per  mil-foot  of  annealed  copper  is  10.4 
ohms  at  20°  C, 


78         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Example  2.  What  is  the  resistance  per  mile  at  20°  G.  of 
an  annealed-copper  conductor  J  inch  in  diameter? 

I  in.  =  0.125  in.  =  125  mils 

7?      pl 

R==~d* 

_  10.4  X  5280 
125  X  125 

=  3.51  ohms 

33.  Resistivity  of  Metals  used  as  Electrical  Conductors. 

The  three  metals  most  used  as  electrical  conductors  are 
copper,  aluminum  and  steel. 

Soft-drawn  copper  has  a  resistivity  of  1.724  microhm- 
centimeters  at  20°  C.  and  is  the  metal  in  greatest  use  as  a 
conductor. 

Hard-drawn  copper  is  nearly  50%  greater  in  tensile 
strength  and  has  a  resistivity  of  1.772  microhm-centimeters, 
less  than  3%  greater  than  annealed  copper,  and  is  therefore 
extensively  used  for  transmission  lines. 

Aluminum  has  a  resistivity  of  2.828  microhm-centimeters 
at  20°  C.,  about  1.6  that  of  copper,  but  this  disadvantage  is 
often  more  than  offset  by  its  small  weight  per  cubic  centi- 
meter, a  conductor  of  aluminum  having  less  resistance  than 
a  conductor  of  copper  of  the  same  length  and  weight. 

The  resistivity  of  steel  depends  largely  upon  its  compo- 
sition and  treatment,  but  it  is  always  much  greater  than 
that  of  copper  or  aluminum.  It  is  approximately  21.6 
microhm-centimeters  for  steel  wire  and  varying  from  13.8 
to  21.6  for  steel  rails. 

The  resistivity  of  alloys  is  almost  always  higher  than  that 
of  any  one  of  the  constituent  metals.  Some  alloys  of  copper, 
nickel,  zinc,  manganese  and  chromium  used  as  resistors  have 
very  high  resistivities  often  approaching  100  microhm- 
centimeters.* 

*  See  Appendix  for  Resistivity  Tables. 


THE  COMPUTATION  OF  RESISTANCE  79 

Prob.  2-4.  What  line  drop  is  there  in  a  2-mile  hard-drawn 
copper  trolley  wire  carrying  200  amperes  if  the  wire  is  0.625 
inch  in  diameter? 

Prob.  3-4.  Each  lamp  in  Fig.  42  takes  2  amperes  at  112 
volts.  The  lamps  are  500  feet  from  the  generator.  The  line 
wire  is  annealed  copper  i  inch  in  diameter.  What  is  the  voltage 
of  the  generator? 


FIG.  42.    A  bank  of  lamps  fed  by  the  generator  G. 


Prob.  4-4.  What  will  be  the  drop  per  mile  in  a  line  consist- 
ing of  annealed-copper  wire  &  inch  in  diameter  carrying  15 
amperes? 

Prob.  5-4.  What  will  be  the  line  drop  in  voltage  and  the  loss 
in  watts  per  mile  in  transmitting  12  kilowatts  at  550  volts, 
if  an  annealed-copper  wire  is  used  having  a  diameter  of  0.364 
inch? 

Prob.  6-4.  A  group  of  incandescent  lamps  takes  12  amperes. 
The  line  drop  is  not  to  exceed  2.6  volts.  What  must  be  the 
size  of  annealed-copper  wire  to  be  used  if  the  lamps  are  2500 
feet  from  the  generator? 

34.  Conductivity  of  Materials.  In  specifying  the  grade 
of  material  to  be  used  for  a  conductor,  it  is  customary  to 
specify  the  conductivity  instead  of  the  resistivity.  Just 
as  conductance  is  the  reciprocal  of  resistance,  so  conductivity 
is  the  reciprocal  of  resistivity. 


(3) 


in  which 


p  =  the  resistivity  in  any  system  of  units, 

7  =  the  conductivity  in  the  same  system  of  units. 


80         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Thus  the  conductivity  of  soft-drawn  copper  at  20°  C.  is 

7  =  0.000001724  =  fi80'000  mh°-<*nti 
The  conductance  of  a  wire  may  be  computed  from 

<?=77'  (4) 

where  G  is  the  conductance  in  mhos  and  the  other  quantities 
have  the  same  meanings  as  before. 

Another  term  in  common  use  is  "  percentage  conductivity." 
By  this  is  meant  the  percentage  which  the  conductivity  of  a 
certain  material  is  of  the  conductivity  of  Standard  Annealed 
Copper. 

Standard  Annealed  Copper  at  20°  C.  has  a  resistivity  of 
1.7241  microhm-centimeters  and  a  density  of  8.89  grams  per 
cubic  centimeter.  The  conductivity  of  this  standard  is 
called  100%.  The  percentage  conductivity  of  any  material 
is  rated  as  a  certain  percentage  of  this  standard.  Thus  a 
rating  of  95%  conductivity  for  a  material  means  that  the 
material  has  a  conductivity  which  is  95%  of  that  of  stand- 
ard copper,  or  0.95  X  580,000  mho-centimeters  conductivity. 
The  resistivity  of  such  a  material  is  1.724/0.95  microhm- 
centimeters. 

Ordinary  copper  runs  between  98%  and  100%  conductivity, 
but  copper  may  be  obtained  which  is  purer  than  the  Stand- 
ard and  has  more  than  100%  conductivity. 

Aluminum  averages  61%  conductivity. 

Prob.  7-4.  What  is  the  resistance  per  mil-foot  of  a  lot  of 
copper  having  96  %  conductivity? 

Prob.  8-4.  What  percent  conductivity  has  a  solid  round 
aluminum  wire  0.365  inch  in  diameter,  one  mile  of  which  has  a 
resistance  of  0.672  ohm  at  20°  C.? 

Prob.  9-4.  What  diameter  must  a  copper  wire  of  96% 
conductivity  have  if  a  mile  of  it  is  to  have  the  same  resistance  as 
the  aluminum  wire  in  Prob.  8-4? 

35.  Temperature  Coefficient  of  Resistance.  It  will  be 
noticed  that  when  the  resistance  of  a  mil-foot  of  copper  wire 


THE  COMPUTATION  OF  RESISTANCE  81 

was  given  as  10.4  ohms  and  that  of  aluminum  as  17.1  ohms, 
the  metal  was  assumed  to  be  at  a  temperature  of  20°  C. 
The  reason  for  stating  the  temperature  is  that  experience 
shows  that  the  resistance  of  any  pure  metal  changes  with 
the  temperature.  For  each  degree  that  the  temperature  of 
a  copper  wire  rises  above  20°  C.  (up  to  about  200°  C.),  the 
resistance  increases  0.393  of  1  %  of  what  it  was  at  20°  C. 
Similarly  for  each  degree  that  the  temperature  of  a  copper 
wire  falls  below  20°  C.  (down  to  about  -  50°  C.),  the  re- 
sistance decreases  0.393  of  1  %  of  what  it  was  at  20°  C. 
This  percentage  change  in  resistance  is  called  the  Tem- 
perature Coefficient  of  Resistance.  For  all  pure  metals  this 
coefficient  has  nearly  the  same  value.*  Since  the  resistance 
per  mil-foot  of  copper  has  been  given  as  10.4  ohms  at  20°  C., 
all  computations  of  resistance  of  wires  based  on  this  value 
will  give  the  resistances  at  20°  C.  In  order  to  find  the 
resistance  of  a  wire  at  any  other  temperature,  it  is  necessary 
to  find  the  increase  or  decrease  in  resistance  and  add  it  to 
or  subtract  it  from  the  resistance  at  20°  C. 

Example  3.     The  resistance  of  a  coil  of  copper  wire  at  20°  C. 
is  48  ohms.     What  will  be  the  resistance  of  the  coil  at  50°  C.? 
The  temperature  rise  =  50°  -  20°  =  30°. 
For  every  degree  rise,  the  resistance  increases  0.393%. 
For  30°  rise,  the  resistance  of  the  coil  increases  30  X  0.393  = 

11.79%. 
The  increase  in  resistance  =  11.79%  of  48  ohms, 

=  5.66  ohms. 
The  resistance  at  50°          =48+5.66 

=  53.66  ohms. 

The  process  used  in  the  above  example  may  be  expressed 
by  an  equation 

#2  =  Ri  [l  +  ai(fe-*i)L  (5) 

in  which  R  i  =  the  resistance  at  the  temperature  t\, 
Rz  =  the  resistance  at  temperature  fe, 
ai  =  the  temperature  coefficient  of  resistance  for 
the  material  at  temperature  ti. 
*  See  Appendix  Table  I 


82         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

We  have  seen  that  on  for  copper  at  20°  C.  is  0.00393. 
For  any  other  temperature,  (fe),  the  value  («2)  of  the  tem- 
perature coefficient  may  be  found  from  the  equation 


Example  4.     What  is  the  temperature  coefficient  of  resistance 
of  annealed  copper  at  30°  C.? 


«1 


1  +  «i  (Tt  -  TJ 

=  0.00393 

"  1  +  0.00393  (30  -  20) 

=  0.00378 

Prob.  10-4.  Prove  that  the  above  equation  for  «2  is  correct 
and  explain  how  the  coefficient  changes  with  a  change  of  the 
temperature  used  as  a  base,  although  the  resistance  change  per 
degree  is  constant. 

Prob.  11-4.  The  resistance  of  a  copper  wire  is  4.90  ohms 
at  20°  C.  What  is  it  at  80°  C.? 

Prob.  12-4.  The  resistance  of  the  field  coils  of  a  generator 
is  220  ohms  at  20°  C.  When  the  coils  become  heated  to  75°  C. 
what  will  the  resistance  be? 

Prob.  13-4.  What  will  the  resistance  of  a  coil  of  copper 
wire  become  at  7°  C.,  if  the  resistance  is  200  ohms  at  20°  C.? 

Prob.  14-4.  What  will  the  resistance  of  a  coil  of  copper 
wire  become  at  7°  C.,  if  the  resistance  is  200  ohms  at  0°  C.? 

Prob.  15-4.  The  resistance  of  a  field  coil  is  130  ohms  at 
12°  C.  What  will  it  be  at  180°  C.? 

Prob.  16-4.  What  will  be  the  resistance  of  a  copper  wire  at 
10°  C.,  if  the  resistance  at  45°  C.  is  2.08  ohms? 

36.  Temperature  Change  Measured  by  Change  in  Re- 
sistance. Electrical  machines  are  generally  sold  under  a 
guarantee  that  the  wire  in  the  coils  will  not  rise  more  than 
a  given  number  of  degrees  when  running  under  a  specified 
load  for  a  specified  time. 


THE  COMPUTATION  OF  RESISTANCE 


83 


By  measuring  the  resistance  of  the  coils  when  at  room 
temperature  (20°  C.  or  68°  F.)  and  then  again  at  the  close 
of  the  run  and  applying  the  equation  for  temperature  effect, 
the  average  temperature  rise  can  easily  be  found. 

Example  5.  The  primary  coils  of  a  transformer  have  a 
resistance  of  5.48  ohms  at  20°  C.  After  a  run  of  2  hours,  the 
resistance  has  risen  to  6.32  ohms.  What  is  the  temperature 
rise  of  the  coil? 

The  resistance  increase  =  6.32  —  5.48  -=  0.84  ohm. 

f\  Qyf 

The  percentage  increase  =  7-75  —  15.3  %. 

5.4o 

The  percentage  increase  for  1°  rise  above  20°  C.  =  0.393% 
To  produce  15.3%  the  temperature  rise  must  be 


15.3 
0.393 


=  38.9°. 


The  average  temperature  rise  in  the  coil  =  38.9°. 


FIG.  43.  A  curve  showing  the  relation  between  resistance  and 
temperature  of  copper. 

Instead  of  making  the  computations  of  the  above  example, 
it  is  generally  more  convenient  to  use  the  equations  of  the 


84         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

graph  showing  the  relation  between  temperature  and  re- 
sistance. Fig.  43  shows  this  graph  for  Standard  Annealed 
Copper.  For  other  pure  metals  the  slopes  would  vary  some- 
what and  the  intercepts  on  the  horizontal  axis  would  differ 
slightly  from  —234.5°.  The  equation  of  this  curve  is 

R*  '    234.5  +  k 


&      234.5  +  U 

Example  6.     The  above  example  would  be  worked  out  ae 
follows  by  this  equation. 

*  '•'  -  234.5 


=  59.1°. 

U  -  *i  =  59.1°  -  20° 
=  39.1°  C.  rise. 

Prob.  17-4.  The  cold  (20°  C.)  resistance  of  an  armature 
was  2.18  ohms.  The  hot  resistance  was  2.56  ohms.  What  was 
the  temperature  rise? 

Prob.  18-4.  It  is  generally  specified  that  the  temperature 
of  the  field  coils  of  a  dynamo  must  not  rise  more  than  65°  on 
full  load.  The  resistance  of  a  set  of  field  coils  before  running 
was  80  ohms  at  20°  C.  After  run  of  3  hours  at  rated  load  the 
resistance  became  92.4  ohms.  Did  the  machine  meet  the  usual 
specifications? 

Prob.  19-4.  The  resistance  of  certain  coils  in  a  machine 
was  found  to  be  7.46  ohms  at  a  temperature  of  40°.  It  was 
specified  that  if  the  machine  ran  continuously  under  full  load, 
the  temperature  of  these  coils  should  not  exceed  105°  C.  (the 
usual  limit).  The  resistance  of  the  coils,  measured  after  a  long 
full-load  run,  was  found  to  be  9.59  ohms.  Did  the  machine 
meet  this  specification? 

37.  Temperature  Coefficient  of  Alloys,  etc.  It  has  been 
stated  that  the  temperature  coefficients  of  resistance  for  all 
pure  metals  are  nearly  the  same,  that  is,  somewhere  about 
0.4%.  Alloys,  although  in  general  of  a  much  higher  re- 


THE  COMPUTATION  OF  RESISTANCE  85 

sistance  per  mil-foot,  have  much  lower  coefficients,  some 
having  zero  and  even  negative  coefficients  at  certain  tem- 
peratures. 

"  Manganin,"  an  alloy  consisting  of  copper,  nickel,  iron 
and  manganese,  for  instance,  has  a  resistance  per  mil-foot 
of  from  250  to  450  ohms,  according  to  the  proportions  of  the 
different  metals  used,  and  a  temperature  coefficient  so  low 
as  to  be  practically  neg  igible.  Manganin  wire  is  hence 
used  to  wind  instruments  where  a  fixed  resistance  is  necessary, 
as  in  the  resistance  coils  of  a  bridge  or  potentiometer.  Before 
the  war  all  manganin  came  from  Germany.  It  is  now  made 
successfully  in  this  country. 

Certain  substances,  notably  carbon,  porcelain  and  glass, 
have  large  negative  temperature  coefficients  of  resistance 
and  decrease  in  resistance  rapidly  when  heated.  The  cold 
resistance  of  a  carbon  lamp  filament  is  about  twice  as  great 
as  the  hot  resistance.  The  porcelain  "  glower  "  of  a  Nernst 
lamp  when  cold  is  a  very  poor  conductor  but  when  heated 
to  incandescence  it  becomes  a  fairly  good  conductor.  The 
filaments  of  tungsten  lamps  are  pure  metal  and  accordingly 
have  a  positive  coefficient  which  is  about  0.0051  at  low  tem- 
peratures. 

38.  Copper-Wire  Tables.  Tables  have  been  prepared 
by  the  Bureau  of  Standards  and  adopted  by  the  A.  I.  E.  E. 
which  give  the  resistance  of  1000  feet  of  standard  annealed- 
copper  wire  of  different  standard  sizes  and  several  tempera- 
tures. The  sizes  are  designated  by  gauge  numbers,  the 
diameter  in  mils  and  the  section  area  in  circular  mils,  etc. 
There  are  several  standard  wire  gauges.  Brown  &  Sharpe 
(B.  &  S.)  is  in  general  use  in  America,  and  is  commonly 
called  the  "American  Wire  Gauge"  (A.  W.  G.).  The 
Birmingham  Wire  Gauge  (B.  W.  G.)  is  in  general  use  in 
Great  Britain.  Table  No.  1  gives  the  complete  data  for 
the  American  or  Brown-&-Sharpe  gauge  numbers.  By 
means  of  these  tables  it  is  easy  to  find  the  resistance  of  any 
length  of  wire  of  a  given  section  area,  etc. 


86        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Example  7.  What  copper  wire  (B.  &  S.  gauge)  should  be 
used  to  transmit  electric  power  2  miles  (out  and  back)  if  the 
resistance  is  not  to  exceed  2.7  ohms  and  the  temperature  to  be 
assumed  is  20°  C.? 


2  miles  =  2  X  5280  =  10,560  feet. 

2.7 

1056  °hmp 
0.256  ohm  per  thousand  feet. 


2.7 

2.7  ohms  for  2  miles    =      '     ohm  per  thousand  feet 
lU.ou 


From  wire  table: 

No.  5  =  0.3133  ohm  per  thousand  feet, 
No.  4  =  0.2485  ohm  per  thousand  feet. 

No.  4  must  be  used  in  order  not  to  exceed  the  limit  of  0.256  ohm 
per  thousand  feet. 

If  the  following  simple  facts  concerning  the  above  table 
are  memorized,  the  gauge  number  and  resistance  and  size 
of  any  wire  can  be  roughly  estimated  without  reference  to 
the  table. 

No.  10  wire  is  practically  ^  inch  (100  mils)  in  diameter, 
or  10,000  circular  mils  area  and  has  practically  1  ohm  re- 
sistance per  1000  feet. 

As  the  wires  grow  smaller,  every  third  gauge  number 
halves  the  section  area  and  doubles  the  resistance.  For 
instance,  No.  13  has  about  5000  circular  mils  area  and  2 
ohms  resistance  per  1000  feet;  No.  16  has  2500  circular 
mils  area  and  4  ohms  per  1000  feet,  etc.  As  the  wires  in- 
crease in  size,  every  third  gauge  number  doubles  the  circular 
mils  area  and  halves  the  resistance;  No.  7  for  instance,  has 
practically  20,000  circular  mils  and  0.5  ohm  per  1000 
feet,  etc. 

Another  simple  method  for  remembering  the  approximate 
resistances  and  weights  of  the  different  gauge  sizes  of  copper 
wires  is  given  in  circular  No.  31  of  the  Bureau  of  Stand- 
ards. 


THE  COMPUTATION  OF  RESISTANCE  87 

Gauge  Number  Ohms  per  1000  Feet 

0 0.1 

1 0.125 

2 0.16 

3 0.2 

4 0.25 

5 0.32 

6 0.4 

7 0.50 

8 0.64 

9 0.8 

10 1. 

11 1.25 

12 1.6 

20. .  10. 

21 12.5 

22 16. 

Note  that  the  resistance  of  No.  0  is  0.1  ohm,  No.  1  is  0.125 
ohm  and  No.  2  is  0.16  ohm.  The  next  number  in  each  col- 
umn is  the  third  gauge  number  and  the  resistance  is  doubled 
in  each  case.  Similarly  for  gauge  Numbers  10,  11  and  12, 
the  resistances  are  just  10  times  those  for  No.  0,  1  and  2, 
respectively,  etc.  It  is  merely  necessary  to  remember  that 
for  gauges  No.  0,  1  and  2  the  resistances  are  0.1,  0.125  and 
0.16  and  that  for  every  third  number  the  resistance  doubles. 

The  weight  may  be  approximated  from  the  fact  that 
1000  feet  of  No.  0  weigh  approximately  320  pounds,  and 
that  the  weight  varies  inversely  with  the  resistance. 

Example  8.  If  the  line  in  Example  7  is  to  work  at  a 
temperature  of  45°  C.,  what  number  wire  will  be  required? 

Solution.  The  resistance  is  not  given  in  the  tables  at  45°  C., 
but  at  20°  C.  We  must  then  find  out  what  resistance  a  wire 
will  have  at  20°  C.  if  it  has  2.7  ohms  resistance  at  45°  C.  The 
temperature  rise  above  20°  is  45  -  20  or  25  degrees.  For  each 
degree  rise  the  resistance  has  increased  0.393%.  For  25  de- 
grees rise  the  resistance  has  increased  25  X  0.393  =  9.825%. 


88        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Thus  at  45°  the  resistance  is  109.83%  of  what  it  was  at  20°. 
The  resistance  at  20°  therefore  equals 


The  resistance  per  1000  feet     = 


109.83 

=    2.48  ohms  for  the  2  miles. 

2.48 


10.56 
0.235  ohm. 


The  problem  thus  becomes:     What  size  of  wire  has  a  resistance 
of  0.235  ohm  per  1000  feet? 
From  table  (No.  5), 

No.  4  has  a  resistance  of  0.2485  ohm  per  1000  feet, 

No.  3  has  a  resistance  of  0.1970  ohm  per  1000  feet. 
No.  3,  therefore,  must  be  used  in  order  not  to  exceed  0.235  ohm 
per  1000  feet. 

Prob.  20-4.  What  size  wire  (B.  &  S.)  will  give  a  resistance 
of  practically  1  ohm  for  the  circuit  of  the  example  above  at 
35°  C.? 

Prob.  21-4.  Thirty-five  50- watt,  110- volt  lamps  are  to  be 
used  in  a  building  so  situated  that  it  requires  200  feet  of  feeder 
wires  (each  way)  from  the  generator  to  the  distributing  point. 
What  size  wire  should  be  run  in  order  that  there  shall  not  be 
more  than  a  3- volt  drop  in  the  feeders? 

Prob.  22-4.  How  far  can  20  amperes  be  transmitted  through 
a  No.  6  wire  (B.  &.  S)  with  4  volts  line  drop? 

Prob.  23-4.  A  coil  for  an  electromagnet  has  800  turns  of 
No.  23  (B.  &  S.)  copper  wire.  The  average  length  of  a  turn  is 
14  inches.  What  is  the  resistance  of  the  coil? 

Prob.  24-4.  It  is  desired  to  construct  a  coil  of  not  more  than 
290  ohms  resistance.  The  coil  must  have  200  turns  of  about 
16  inches  average  length.  What  size  wire  (B.  &  S.)  should  be 
used? 

39.  Stranded  Wire.  On  account  of  their  greater  flexi- 
bility, stranded  cables  are  often  used  instead  of  solid  wires. 
A  stranded  cable  is  much  easier  to  pull  into  a  conduit  and 
less  likely  to  break  when  bent  at  sharp  angles.  When 
a  size  of  wire  larger  than  No.  0000  is  required,  it  is  prae- 


THE  COMPUTATION  OF  RESISTANCE  89 

tically  always  made  in  strands  rather  than  solid  but  the 
smaller  sizes  are  also  very  common  in  the  stranded  form. 

For  instance,  instead  of  using  a  solid  No.  4  wire,  having 
a  diameter  of  204  mils  and  an  area  of  41,700  circular  mils, 
it  is  much  easier  to  use  a  cable  made  up  of  7  wires  each 
0.077  inch  in  diameter.  Each  strand  (wire)  would  then 
have  an  area  of  77  X  77,  or  5930  circular  mils.  Since  the 
cable  is  made  up  of  7  of  these  strands,  the  area  of  the  cable 
would  be  7  X  5930  or  41,500  circular  mils,  which  is  practi- 
cally the  area  of  a  No.  4  solid  wire.  The  diameter  of  a 
stranded  wire  will  always  be  slightly  greater  than  that  of  a 
solid  wire  of  an  equivalent  cross-section. 

On  the  other  hand,  due  to  the  spiral  arrangement  of  the 
strands,  the  effective  length  as  well  as  the  mass  is  increased 
beyond  what  it  would  be  for  a  rod  of  the  same  cross-section 
(number  of  strands  times  cross-section  of  each  strand).  The 
resistance  is  also  increased.  It  is  therefore  customary  to 
compute  the  resistance  of  a  stranded  cable  by  adding  2% 
to  the  resistance  of  a  rod  of  the  same  cross-section.  This 
2%  correction  holds  for  only  one  value  of  the  "  lay  "  of  the 
strands.  For  the  method  of  computing  the  correction,  see 
U.  S.  Bureau  of  Standards  Circular,  No.  31. 

Prob.  25-4.  To  what  size  wire  (B.  &  S.)  is  a  stranded  cable 
equivalent  which  is  made  up  of  19  strands  each  0.059  inch  in 
diameter? 

Prob.  26-4.  It  is  desired  to  make  a  cable  of  19  strands 
which  shall  be  equivalent  to  a  No.  0  (B.  &  S.)  solid  conductor. 
What  size  strands  should  be  used? 

Prob.  27-4.  How  many  strands  0.061  inch  in  diameter  will 
it  take  to  make  a  cable  equivalent  to  a  No.  6  wire? 

Prob.  28-4.  It  is  desired  to  make  a  very  flexible  cable 
equivalent  to  No.  4  (B.  &  S.)  wire.  If  strands  of  No.  22,  (B.  & 
S.)  wire  are  used,  how  many  will  be  required? 

40.  Aluminum.  We  have  noted  that  although  the  re- 
sistance of  aluminum  wire  is  17.1  ohms  per  mil-foot  (prac- 
tically 1.6  times  that  of  copper)  its  weight  is  only  0.3  that 


90         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

of  copper.  For  this  reason  some  transmission  lines  are 
strung  with  aluminum  wire.  While  this  necessitates  a 
larger  wire  for  the  same  resistance  per  1000  feet,  the  weight 
of  such  a  wire  will  be  less  than  that  of  a  copper  conductor. 
However,  such  a  line  possesses  a  disadvantage  over  a  cop- 
per line  in  that  aluminum  melts  at  a  lower  temperature 
than  copper.  A  short  circuit  which  would  burn  up  but  a 
few  inches  of  a  copper  wire  is  likely  to  burn  out  long  sections 
of  an  aluminum  line.  Likewise  having  a  greater  cross- 
section  it  collects  a  greater  weight  of  ice  per  foot  during  a 
sleet  storm.  This  is  likely  to  weight  the  wire  beyond  its 
tensile  strength.  Aluminum  is  also  more  difficult  to  splice 
satisfactorily  than  copper,  which  adds  to  the  cost  of  con- 
struction of  a  line  with  aluminum  conductors. 

Example  9.  What  size  would  an  aluminum  wire  be  which 
has  the  same  resistance  as  a  No.  4  copper  wire? 

Solution.  No.  4  copper  wire  has  a  resistance  per  1000  feet 
of  0.2485  ohm.  The  resistance  of  an  aluminum  wire  is  found 
from  the  equation 

7?        l7'1 
R  '    ~* 


Thus,  0.2485  •  17'1  *  1Q°° 


=  17.1  X  1000 

0.2485 
=  68,800  circular  mils. 

This  would  require  an  aluminum  wire  of  gauge  No.  1,  of  83,690 
circular  mils.  No.  2  has  a  cross-section  of  only  66,370  circular 
mils.  An  aluminum  wire  of  this  size  would  be  called  the  equiv- 
alent of  No.  4  copper  wire. 

Example  10.  How  would  the  weight  per  1000  feet  of  the 
equivalent  aluminum  wire  in  the  above  example  compare  with 
the  weight  of  the  copper  wire? 

The  weight  of  1000  ft.  of  No.  4  copper  wire  =  126  Ib. 
The  weight  of  1000  ft.  of  68,800  cir.  mil  copper  wire  =  208.2  Ib. 
The  weight  of  1000  ft.  of  68,800  cir.  mil  aluminum  wire  =  0.30 

X  208.2  =  62.5  Ib. 


THE  COMPUTATION  OF  RESISTANCE  91 

Thus  the  weight  of  the  aluminum  equivalent  is  only 

62.5 
126 

or  49.6%  of  that  of  the  copper  wire. 

Prob.  29-4.  What  size  aluminum  wire  will  have  the  same 
resistance  per  mile  as  a  No.  6  (B.  &  S.)  copper  wire? 

Prob.  30-4.  An  aluminum  cable  is  to  be  made  up  of  19 
strands  which  will  be  equivalent  to  a  No.  0  solid  copper  wire. 
What  size  must  the  strands  be? 

Prob.  31-4.  If  83  No.  19  aluminum  wires  are  used  in  making 
up  a  cable,  to  what  size  solid  copper  is  such  a  cable  equivalent? 

41.  Copper-Clad  Steel  Wire.  On  account  of  its  lower 
cost  and  great  tensile  strength,  copper-clad  steel  wire  has 
lately  come  into  use  for  trolley  wires  and  transmission  lines. 
This  type  of  wire  consists  of  a  steel  core  to  which  has  been 
welded  a  covering  of  copper.  The  resistance  per  mil-foot 
of  such  wire  depends  upon  the  relative  sizes  of  the  copper 
and  steel  cross-section  areas.  One  company  has  put  on  the 
market  two  grades,  one  of  so-called  30  %  conductivity  and 
the  other  of  40  %  conductivity.  This  rating  merely  means 
that  copper  wires  would  have  30%  and  40%  respectively 
of  the  resistance  of  the  copper-clad  steel  wire  of  the  same 
size. 

In  the  30%- conductivity  wire  the  area  of  the  steel  core  is 
79.5%  of  the  entire  cross-section  while  the  copper  has  20.5% 
area.  In  the  40%-conductivity  wire,  the  steel  makes  up 
68.2%  and  copper  31.8%  of  the  total  area. 

Prob.  32-4.  In  a  No.  00  copper-clad  trolley  wire  of  30  % 
conductivity,  how  many  circular  mils  would  there  be  of  steel 
and  copper  respectively? 

Prob.  33-4.  What  would  be  the  resistance  of  1000  feet  of 
the  steel  core  of  the  trolley  wire  in  Prob.  32-4?  The  mil- 
foot  resistance  of  steel  =  86.7  ohms. 

Prob.  34-4.  What  would  be  the  resistance  of  1000  feet  of 
the  copper  cover  of  the  trolley  wire  of  Prob.  32-4? 


92         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  35-4.  From  the  resistance  of  the  steel  core  and  copper 
covering  of  1000  feet  of  the  trolley  wire  in  Prob.  32-4,  com- 
pute the  resistance  of  the  trolley  wire.  Would  1000  feet  of 
a  copper  wire  of  the  same  size  have  30%  as  much  resistance? 

42.  Safe  Carrying  Capacity  for  Copper  Wires.     In  in- 
stalling wire  in  buildings,  it  is  necessary  to  take  into  account 
another  factor  besides  the  voltage  drop  when  determining 
the  size  to  be  used.     An  electric  current  heats  any  conductor 
through  which  it  passes.     If  heat  is  generated  in  the  wire 
faster  than  it  can  be  dissipated  from  the  surface  of  the  wire, 
the  temperature  will  continue  to  rise  as  long  as  this  condition 
exists.     It  is  necessary  therefore  to  select  a  wire  which  will 
dissipate  the  heat  generated  by  the  current  at  such  a  rate 
that  the  temperature  will  never  rise  high  enough  to  cause  the 
insulation   to   deteriorate.     The   National   Board   of   Fire 
Underwriters  has  therefore  issued  a  table  of  the  safe  current 
capacity  of  copper  wire  of  the  sizes  used  in  house  wiring. 
Wherever  local  regulations  do  not  specify  otherwise,  the 
currents  carried  by  any  interior  wiring  should  not  exceed 
the  values  given  in  this  table.     See  Appendix. 

Example  11.  It  is  desired  to  install  a  conductor  to  carry 
40  amperes.  What  size  copper  wire  should  be  used? 

From  the  table,  No.  6  rubber-insulated  wire  will  carry  46 
amperes  and  is  the  size  to  be  used. 

If  weather-proof  wire  can  be  used,  No.  8  will  do. 

43.  Determination  of  Right  Sizes  for  Interior  Wiring. 
In  deciding  upon  the  sizes  which  should  be  used  in  the  dif- 
ferent parts  of  any  interior  distributing  system,  it  is  neces- 
sary to  take  into  consideration  two  factors,  — 

First:  The  size  in  each  section  must  be  such  that  the 
current  in  no  wire  exceeds  the  amount  given  in  the  Under- 
writers' Table  of  safe  carrying  capacities  for  wires.  It  is 
therefore  necessary  to  determine  accurately  the  current 
which  each  wire  must  carry  and  to  make  a  tentative  selec- 
tion of  size  from  the  above  table. 


THE  COMPUTATION  OF  RESISTANCE 


93 


Second:  The  voltage  drop  throughout  the  system  must 
then  be  computed  in  order  to  make  certain  that  it  does  not 
exceed  certain  values,  for  if  lamps  are  to  be  operated  any- 
where on  the  system  a  variation  of  more  than  5%  in  the 
voltage  at  the  lamps  causes  an  unpleasant  variation  in  the 
illumination.  If  the  entire  load  consists  of  motors,  heating 
appliances,  etc.,  a  drop  of  10%  is  usually  allowable.  Any 
greater  drop  than  this,  however,  would  have  a  bad  effect 
upon  the  speed  of  the  motors. 

Example  12.  The  panel  board  (P.B.)  in  Fig.  44  is  situated 
150  feet  from  the  main  switch.  From  the  board  run  three 


I 

rCh 
O 
0 
-0- 

<|>6$<J>   . 

^>OPQ 

tpR 
P.B. 

{^   5 

to*M 

i 

0 
•0 
-0 

L<>J 

£533 

r^666 

i 

FIG.  44.    A  distribution  diagram  for  house  wiring. 

branch  lines.     Each  branch  is  supplied  with  12  outlets  for  50- 
watt,  110- volt  lamps.     What  size  must  the  mains  be? 

Solution.     Each  branch  line  carries   12  X  50  or  600  watts. 


600 
This  means  — —  or  5.45  amperes  in  each  branch  wire. 


Branch- 


wire  sizes  are  determined  according  to  Table  III.  Although 
from  the  table  it  is  observed  that  No.  16  wire  would  carry  this 
current  safely,  it  will  be  noted  that  no  size  smaller  than  No.  14 
can  be  installed  in  order  that  the  wires  may  have  sufficient 
mechanical  strength. 
Mains.  Size  according  to  Table  III. 

Each  main  must  carry  the  current  in  all  three  branches  or 
3  X  5.45  =  16.35  amperes. 

According  to  Table  III,  No.  12  wire  must  be  used  as  the  next 
size  in  common  use,  No.  14,  can  carry  only  15  amperes. 
Checks  on  above  sizes  for  voltage  drop. 


94         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

The  distance  from  the  panel  to  the  load  center* 

of  branches  =     50  ft. 

The  length  of  wire  in  each  branch  =  2  X  50  =100  ft. 

The  resistance  of  1000  ft.  No.  14  (Table  II)  =  2.525 

The  resistance  of  100  ft.  of  No.  14  =  TV  of  2.525  =  0.253  ohm 
The  voltage  drop  in  each  branch    =   5.45  X 

0.253  =  1.38  volts 

The  length  of  the  mains  =  2  X  150  ft.  =  300  ft. 

The  resistance  of  1000  ft.  of  No.  12  (Table  II)  =  1.588  ohms 
The  resistance  of  300  ft.  of  No.  12  =  0.3  of 

1.588  =  0.476  ohm 

The  voltage  drop  in  the  mains  =  16.35  X  0.476  =  7.78    volts 
The  total  drop  from  the  main  switch  to  the 

lamps  =  7.78  +  1.38  =  9.16  volts 

9  16 
The  percentage  line  drop  =  -^— -  =  8.3% 

This  is  nearly  twice  as  great  a  drop  as  is  allowed  in  good 
practice  because  the  brightness  of  the  lamps  would  vary  through 
wide  ranges  depending  on  how  many  were  in  use  at  one  time. 
When  only  a  few  lamps  were  in  use,  the  voltage  of  these  lamps 
would  be  about  the  same  as  that  at  the  main  switch,  110  +  9.16, 
or  about  119  volts.  The  voltage  at  this  main  switch  would 
have  to  be  119  volts  in  order  to  maintain  110  volts  at  the  lamps 
on  full  load.  This  would  cause  the  lamps  to  glow  far  above  their 
rated  candle  power  and  would  either  burn  them  out  at  once  or 
shorten  their  life  to  a  small  percent  of  the  normal  rating.  It 
would,  therefore,  be  necessary  to  install  larger  than  No.  12 
mains.  Let  us  try  No.  10. 

The  resistance  of  300  ft.  of  No.  10  =  0.3  of  0.9989  =  0.300 
ohm. 

The  drop  in  the  main  =  0.300  X  16.35  =  4.91  volts. 
The  total  drop  =  4.91  +  1.38  =  6.29  volts. 

f*     C\f\ 

The  percentage  drop  =  -1— -  =  5.72%. 

This  is  still  somewhat  too  large  a  drop.     It  is,  therefore,  neces- 
sary to  use  No.  8  mains. 

*  Note:  The  load  center  is  that  point  on  the  branch  line  at  which, 
for  convenience  in  calculation,  all  lamps  may  be  considered  to  be 
concentrated. 


THE  COMPUTATION  OF  RESISTANCE  95 

Prob.  36-4.  What  size  mains  would  be  used  in  the  above 
example  if  the  panel  board  were  situated  50  feet  from  the  main 
switch  and  the  same  loads  were  on  the  branch  lines? 

Prob.  37-4.  An  installation  requires  seventy  50-watt  lamps. 
The  panel  board  is  situated  80  feet  from  the  main  switch. 
What  size  main  should  be  run?  Note  that  the  Underwriter 
Rules  do  not  ordinarily  allow  more  than  twelve  50-watt  lamps 
on  a  single  branch  line. 

Prob.  38-4.  If  the  panel  board  in  the  above  problem  could 
be  placed  40  feet  from  the  main  switch,  what  size  mains  might 
be  used? 

Prob.  39-4.  If  the  load  on  the  installation  of  Prob.  37-4 
consisted  of  motors  instead  of  lamps  but  using  the  same  total 
load,  what  size  wire  could  be  used  for  the  mains? 

44.  Insulating  Materials.  Even  the  materials  which  we 
call  insulators  conduct  electricity  to  a  certain  small  extent. 
The  resistance  of  such  materials  is  very  high  compared  with 
the  resistance  of  metals.  The  resistivity  of  soft  copper  is 
0.00000172  ohm-centimeter,  which  may  be  written  more 
conveniently 

1.72  X  10~6  ohm-centimeter. 

Expressed  in  the  same  way  the  resistivity  of  glass  under 
ordinary  conditions  is  about 

5  X  1016  ohm-centimeters. 

This  means  that  a  cube  of  glass  one  centimeter  on  a  side 
will  offer  a  resistance  of  50,000,000,000,000,000  ohms.  Such 
an  enormous  resistance  can  be  measured  only  with  great 
difficulty.  In  fact,  the  value  varies  so  much  with  different 
specimens  and  conditions  of  measurement  that  it  is  not  known 
accurately. 

Average  values  of  the  resistivities  in  ohm-centimeters 
for  common  insulating  materials  are  about  as  follows : 

Rubber  2  X  1015, 

Impregnated  paper  5  X  1014, 

Varnished  cambric    2  X  1011, 

Glass  5  X  1018, 

Fused  Quartz  1  X  1018. 


96         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Materials  such  as  the  above  do  not  obey  Ohm's  law. 
When  the  voltage  applied  to  a  specimen  is  doubled,  the 
current  will  be  more  than  doubled.  The  resistivity  of  the 
material  depends  upon  the  voltage  used  in  measuring  it. 
More  accurately,  it  depends  upon  the  voltage  gradient,  or 
the  voltage  per  inch  of  thickness,  which  is  applied.  The 
higher  the  voltage  gradient,  the  lower  will  be  the  resistivity. 
The  values  given  in  the  table  above  are  for  low  potential 
gradients  of  about  1000  volts  per  inch. 

Insulating  materials  have  negative  temperature  coeffi- 
cients of  resistance  which  are  usually  large.  As  the  tem- 
perature is  increased,  the  resistance  of  a  material  rapidly 
diminishes.  In  fact,  all  materials  which  can  stand  the 
temperature  are  fairly  good  conductors  at  a  red  heat.  The 
resistivity  of  glass  at  ordinary  temperatures  will  vary  as 
much  as  10%  for  one  degree  centigrade  change  of  tem- 
perature. 

With  insulating  materials,  the  leakage  over  the  surface 
is  often  greater  than  the  conduction  through  the  body  of 
the  material.  With  a  porcelain  transmission-line  insulator, 
by  far  the  greater  part  of  the  leakage  current  from  the  line 
flows  over  the  surface.  This  is  particularly  true  if  the 
surface  is  dusty  or  wet.  It  is  to  lengthen  these  leakage 
paths  and  to  keep  part  of  them  dry  that  an  insulator  is 
made  with  "  petticoats."  The  surface  resistivity  is  given 
as  the  number  of  ohms  resistance  between  opposite  edges 
of  a  piece  of  the  surface  one  centimeter  square.  It  is  an 
extremely  variable  quantity,  depending  upon  the  humidity, 
the  temperature  and  several  other  factors. 

Example  13.  A  glass  busbar  insulator  is  a  cylinder  10  centi- 
meters high  and  of  4  centimeters  diameter.  A  voltage  of  30,000 
is  applied  between  the  busbar  on  top  and  the  support  on  the 
bottom.  The  temperature  is  40°  C.  Take  the  volume  re- 
sistivity at  this  temperature  and  gradient  as  5  X  1011  ohm- 
centimeters  and  the  surface  resistivity  as  2  X  1010  ohm-cen- 
timeters. How  much  current  in  microamperes  leaves  the  bus- 
bar through  the  insulator? 


THE  COMPUTATION  OF  RESISTANCE  97 

The  volume  resistance  is 


=  4  X  1011  ohms. 
The  current  through  the  volume  of  the  glass  is 

-HI 


=  0.07  microampere. 
The  surface  resistance  is 

Rs  =  ps—9 
w 

where  I  is  the  length  and  w  the  width  of  the  surface  path. 


Rs  =  2  X  1010        =  1.6  X  1010ohms. 
4?r 

The  current  over  the  surface  of  the  glass  is 
E          30000 


=  2  microamperes. 

The  total  current  is  hence  practically  2  microamperes,  that 
conducted  through  the  volume  of  the  glass  being  negligible. 

It  is,  of  course,  useless  to  compute  such  a  problem  with 
great  accuracy,  for  the  constants  of  the  material  are  never 
accurately  known. 

It  will  be  noticed  that  the  amount  of  current  which  leaks 
from  the  wire  in  the  above  example  is  small  compared  with 
the  current  probably  carried  by  the  wire  itself.  This  is 
usually  the  case  in  ordinary  wire  circuits.  Except  in  special 
cases  the  leakage  can  be  entirely  neglected. 

45.  Insulation  Resistance  of  Cables.  The  insulation  re- 
sistance of  electrical  machines,  transformers  or  cables  is 
sometimes  measured  and  is  used  as  a  rough  indication  of 
the  condition  of  the  insulation.  This  is  the  resistance 


98         PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

measured  between  the  copper  wire  and  the  frame  or  sheath. 
A  moderate  size  of  low-voltage  motor,  as  usually  insulated, 
when  warm  from  carrying  full  load,  should  show  an  in- 
sulation resistance  of  about  a  megohm.  When  cold,  the 
insulation  resistance  will  have  a  value  many  times  this 
amount. 

The  insulation  resistance  of  lead-sheathed  cables  is  often 
measured   to   determine  whether  the  insulation  is  deterio- 
rating.    The  insulating  material  of  a  cable,  when  considered 
as  a  conductor  for  leakage  current,  is  not  of  uniform  cross- 
section.     The  resistance  must  hence 
be  found  by  an  integration. 

Fig.  45  shows  a  cross-section  of  a 
single-conductor  cable.  The  flow  of 
leakage  current  is  radial,  out  from 
the  central  copper  conductor  to  the 
outside  lead  sheath.  The  cross- 
section  of  the  path  varies.  We  can 
consider  it  to  be  made  up  of  a 
number  of  elementary  paths  in 
series. 

Consider  a   centimeter  length    of 

the  cable  (perpendicular  to  the  paper).  Map  out  an  element 
of  the  material  of  thickness  dx  and  radius  x.  If  p  is  the 
resistivity  of  the  material,  the  resistance  of  this  element 
will  be 

dR=^. 


A  cross-section 
lead-sheathed 


To  obtain  the  total  resistance  we  add  the  resistances  of  all 
such  elements  between  x  =  r\  and  x  =  r2:  that  is, 

R=    I     s~-  =7^- log  -  ohms  per  centimeter  of  length.    (8) 
«/  7*1    ZirX       &ir       *  T\ 


From  this  formula  we  can  compute  the  insulation  resistance 
of  a  cable. 


THE  COMPUTATION  OF  RESISTANCE  99 

Example  14.  If  p  =  2  X  1014  ohm-centimeters  is  the  re- 
sistivity of  the  insulation  for  a  varnished-cambric  insulated 
cable,  where 

ri  =  1  centimeter 
and 

r2  =  4  centimeters, 

we  have  from  the  above  formula 
2X10" 

R  "   ~~ 


io14 

=  -  2.30  X  Iogi04 

7T 

IO14 
=  --2.30  X0.60 

TT 

=  4.4  X  IO13  ohms  per  centimeter  length  of  cable. 

The  insulation  resistance  of  a  mile  of  this  cable  will  equal 
this  value  divided  by  the  number  of  centimeters  in  a  mile 
(1.61  X  IO5).  The  cable  will  then  have  an  insulation  re- 
sistance of  approximately 

3  X  IO8  ohms  per  mile, 
or 

300  megohms  per  mile 

under  the  above  assumption. 

This  formula  must  be  used  with  considerable  discretion, 
for  insulation  resistance  is  a  variable  matter  depending 
upon  very  many  conditions.  The  derivation  of  the  formula 
will,  however,  show  the  manner  in  which  the  resistance  of  a 
non-uniform  conductor  can  be  computed. 


SUMMARY    OF   CHAPTER  IV 

The  RESISTANCE  of  a  conductor  can  be  computed  from  the 
formula 

R=     -, 

where 

R  =  the  resistance  in  ohms, 
I  =  the  length  of  conductor, 
A  =  the  cross-section  area, 
p  =  the  resistivity  in  units  depending  on  the 
units  of  length  and  cross-section. 

Resistivity  is  generally  measured  in  OHM-CENTIMETERS, 
that  is,  the  resistance  between  parallel  faces  of  a  centimeter 
cube,  or  in  OHMS  PER  MIL-FOOT,  that  is,  the  resistance  of 
a  conductor  one  foot  long  and  one  circular  mil  in  cross-section 
area. 

THE  MATERIALS  MOST  COMMONLY  USED  for  con- 
ductors are  copper  for  its  low  resistivity,  aluminum  for  its  light 
weight  and  steel  for  its  strength. 

CONDUCTIVITY  is  the  reciprocal  of  resistivity.  A  certain 
grade  of  copper  is  taken  as  the  standard  and  is  said  to  have 
100%  conductivity  and  metals  are  rated  as  a  certain  percent 
conductivity  of  that  standard. 

The  TEMPERATURE  COEFFICIENT  OF  RESISTANCE  is 
the  change  in  resistance  per  degree  per  ohm  at  initial  tempera- 
ture. THIS  COEFFICIENT  IS  ABOUT  0.004  for  all  pure 
metals.  For  alloys  it  is  less  and  may  be  zero  or  even  negative. 
For  insulators  it  is  always  negative  and  of  considerably  higher 
order  of  magnitude  than  for  conductors. 

FOR  COPPER,  the  effect  of  temperature  change  may  be 
computed  from  the  equation 

R2      234.6  +  t2 
Ri 


THE  COMPUTATION  OF  RESISTANCE  101 

where 

ti  =  the  initial  temperature, 
RI  =  the  initial  resistance, 

tz  =  the  final  temperature, 
R2  =  the  final  resistance. 

In  the  United  States  the  standard  wire  gauge  is  the  "  Brown 
&  Sharpe "  or  "American  Wire  Gauge."  Copper  wire  is 
commonly  made  in  the  sizes  indicated  in  the  Copper  Wire 
Table  and  has  the  resistance  and  weight  per  1000  feet  indicated. 

STRANDED  WIRE  has  greater  flexibility  but  for  the  same 
cross-section  it  has  2%  more  resistance  per  mile  due  to  the 
spiral  arrangement  of  the  strands. 

ALUMINUM  WIRE  has  about  1.6  the  resistivity  of  copper, 
but  is  only  ^  as  heavy.  It  is  hard  to  splice  and  is  likely  to 
suffer  more  from  ice  storms  and  short  circuits. 

COPPER-CLAD  WIRE  has  a  steel  core  which  lowers  the 
conductivity  but  increases  the  strength. 

THE  SAFE  CURRENT-CARRYING  CAPACITY  OF  WIRES 
is  indicated  in  a  table  issued  by  the  National  Board  of  Fire 
Underwriters.  A  large  current  would  sooner  or  later  injure 
the  insulation. 

THE  CORRECT  SIZE  OF  WIRE  FOR  INTERIOR  WORK 
is  such  that,  first,  no  conductor  will  carry  more  current  than 
that  indicated  in  the  Underwriters  table,  and,  second,  the 
voltage  regulation  of  lamps  shall  not  exceed  6%  and  that  of 
motors  10%. 

THE  SURFACE  LEAKAGE  RESISTANCE  OF  INSUL- 
ATORS is  generally  less  than  the  volume  resistance. 

THE  INSULATION  RESISTANCE  OF  SHEATHED 
CABLES  can  be  computed  from  the  equation 


where 

R  =  the  insulation  resistance  per  centimeter  length 
of  cable, 

p  =  the  resistivity  of  insulation  material, 
r2  =  the  outside  radius  of  insulation, 
ri  =  the  inside  radius  of  insulation. 

The  value  of  p  in  this  equation  depends  upon  many  factors 
and  is  very  uncertain. 


PROBLEMS   ON   CHAPTER   IV 

Prob.  40-4.  How  far  will  a  pair  of  copper  line  wires  trans- 
mit 40  amperes  with  a  line  drop  of  8  volts,  if  the  wire  is  0.262 
inch  in  diameter? 

Prob.  41-4.  What  size  iron  wire  will  have  the  same  re- 
sistance per  mile  as  a  No.  4,  B.  &  S.  copper  wire? 

Prob.  42-4.  To  what  size  copper  wire  is  a  stranded  alu- 
minum cable  equivalent  which  is  made  up  of  19  strands  each 
0.059  inch  in  diameter? 

Prob.  43-4.  How  many  strands  of  aluminum  wire  0.061 
inch  in  diameter  will  it  take  to  make  a  cable  equivalent  to  a 
No.  6  copper  wire? 

Prob.  44-4.  It  is  desired  to  make  a  flexible  aluminum  cable 
equivalent  to  No.  4  copper  wire.  If  strands  of  No.  19,  B.  &  S. 
wire  are  used  how  many  will  be  required?  Cables  are  usually 
made  of  7,  19,  37  or  61  strands.  Use  one  of  these  numbers. 

Prob.  46-4.  What  are  the  cross-sectional  dimensions  of  a 
round  conductor  with  an  iron  core  surrounded  by  copper  having 
a  mean  conductivity  of  40%  at  0°  C.,  and  offering  a  resistance 
of  10  ohms  per  mile  at  25°  C?  The  copper  used  has  a  con- 
ductivity of  100  %,  the  iron  16.8  %,  both  at  0°  C.  The  annealed 
copper  standard  of  10.4  ohms  per  mil-foot  at  20°  C.  is  to  be  taken. 

Prob.  46-4.  It  is  desired  to  determine  the  length  of  wire 
and  the  mass  of  copper  on  the  field  winding  of  a  small  dynamo. 
The  percent  conductivity  of  the  copper  is  not  known,  but  the 
specific  gravity  is  taken  to  be  8.89  and  the  cross-section  of  the 
wire  is  found  by  measurement  to  be  1022  circular  mils.  Ac- 
curate measurements  of  the  resistance  of  the  winding  at  two 
temperatures  are  made  with  the  following  results: 

Resistance  at  20°  C.  =  126.5  ohms, 
Resistance  at  70°  C.  =  150.2  ohms. 

From  the  above  data,  calculate  the  length  of  wire  in  feet, 
the  mass  of  copper  in  pounds  and  the  percent  conductivity 
of  the  copper.  With  your  answer,  state  the  reference  book 
and  page  from  which  you  take  any  necessary  additional  data. 

102 


THE  COMPUTATION  OF  RESISTANCE  103 

Prob.  47-4.  A  transmission  line  of  3  wires,  each  12  miles 
long,  is  to  be  designed  for  a  current  of  110  amperes  in  each 
wire  and  a  total  heat  loss  of  150,000  watts  at  20°  C.  The 
conductors  are  to  be  of  copper  of  97%  conductivity.  What 
is  the  least  weight  of  copper  which  meets  these  conditions? 
What  cross-section  in  circular  mils  would  each  such  conductor 
have?  What  is  the  smallest  A.W.G.  size  of  wire  which  comes 
within  the  conditions  of  the  problem?  If  this  size  of  wire  is 
used  and  has  97%  conductivity,  what  is  the  actual  heat  loss? 

Prob.  48-4.  The  field  coils  of  a  dynamo  are  composed  of 
wire  having  a  mean  temperature  coefficient  of  0.00393  per 
degree  centigrade,  based  on  20°  C.  The  resistance  of  the  coils 
is  116  ohms  after  they  have  long  stood  inert  at  a  temperature 
of  26°  C.  Later  when  heated  by  the  passage  of  the  current, 
their  resistance  is  found  to  be  136.5  ohms.  What  is  their  mean 
temperature  in  the  latter  state? 

Prob.  49-4.  Power  is  to  be  transmitted  over  a  two-wire 
transmission  line  to  a  factory  three  miles  from  the  generating 
station.  The  factory  requires  a  current  of  500  amperes  at  a 
potential  of  600  volts,  and  it  is  specified  that  the  losses  in  the 
transmission  line  shall  not  be  more  than  10%  of  the  received 
energy.  The  resistivity  of  copper  at  20°  C.  is  10.6  ohms  per 
mil-foot  and  that  of  aluminum  17.1  ohms  per  mil-foot.  Their 
temperature  coefficients  at  20°  C.  are  0.00384  and  0.0039  re- 
spectively. If  the  line  is  to  operate  in  a  climate  whose  aver- 
age temperature  is  50°  Fahrenheit,  find  the  smallest  volumes  of 
copper  and  aluminum  it  would  be  possible  to  use  for  the  con- 
struction of  the  entire  line. 

Prob.  60-4.  If  copper  costs  15  cents  per  pound  and  aluminum 
32  cents  per  pound,  which  metal  will  be  the  cheaper  to  use  in 
Prob.  49-4? 

Prob.  51-4.  A  motor  is  connected  to  the  220-volt  busbars 
of  a  power  station  by  means  of  a  one-mile  line  of  aluminum 
wire  (conductivity  60%).  The  cross-section  of  each  wire  is 
33,100  circular  mils.  The  motor  field  is  wound  with  copper 
wire  and  its  resistance  at  20°  C.  is  150  ohms.  The  motor 
armature  is  also  wound  with  copper  and  its  total  resistance  at 
20°  C.  is  0.5  ohm.  Compute  the  resistances  of  all  parts  of  the 
circuit  when  the  temperature  of  the  line  wires  is  30°  C.  and 
of  the  motor  70°  C. 

Prob.  62—4.  Compute  the  weight  of  copper  wire  necessary 
to  replace  the  aluminum  line  wires  in  Prob,  $1-4  and  to  give 
the  same  resistance  at  30°  C. 


104      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  53-4.  Two  industrial  plants,  approximately  2000  feet 
apart,  decide  to  combine  their  power  plants  into  one  situated 
at  one  of  the  plants.  The  generators  are  rated  at  235  volts  and 
the  motors  are  230  volts  and  220  volts  so  divided  in  number 
that  the  230-volt  motors  may  be  used  at  the  plant  near  the 
power  house  and  the  220-volt  motors  at  the  more  distant  plant. 
The  total  demand  of  the  220-volt  motors  is  400  amperes  and 
of  the  230-volt  motors  it  is  700  amperes.  Under  these  condi- 
tions the  voltage  drop  through  the  feeders  of  the  700-ampere 
demand  is  5  volts,  of  the  line  between  the  plant,  10  volts,  and 
of  the  feeders  to  the  other  demands,  5  volts.  The  feeders  in 
each  plant  are  500  feet  long.  Find  the  weight  of  copper  nec- 
essary for  this  arrangement.  The  Electric  Journal,  February, 
1915,  page  7. 

Prob.  54-4.  In  Prob.  53-4  find  the  losses  in  the  line  and 
feeders.  If  the  resistance  of  the  main  switchboard  connections 
and  the  armatures  of  the  machines  in  parallel  is  0.008,  what  is 
the  generated  e.m.f.  of  the  machines  and  what  is  the  loss  in  the 
switchboard  and  armature  windings?  Assume  the  generating 
units  to  be  identical  machines. 

Prob.  55-4.  If  in  Prob.  54-4  one-half  the  motors  are  taken 
out  of  service  at  the  nearer  plant,  what  would  be  the  voltage 
at  the  switchboard,  at  the  other  motors  of  the  other  plant  and 
at  the  motors  of  the  nearer  plant? 

Assume  the  generated  e.m.f.  of  the  machines  to  be  the  same 
as  in  Prob.  53-4. 

Prob.  56-4.  If  in  Prob.  54-4  the  entire  motor  load  is  taken 
off  at  the  more  distant  plant,  what  should  be  the  generated 
voltage  of  the  generators  in  order  to  maintain  230  volts  at  the 
end  of  the  feeder  in  the  nearer  plant?  What  would  be  the  open- 
circuit  voltage  at  the  end  of  the  feeder  in  the  more  distant 
plant? 

Prob.  57-4.  An  aluminum  bar  has  a  uniform  thickness  of 
0.30  inch.  Its  breadth  varies  uniformly  from  9  inches  at 
one  end  to  12  inches  at  the  other.  Its  length  is  18  feet.  Its 
volume  conductivity  is  61  %.  What  is  its  resistance  at  20°  C.? 
What  is  the  percent  error  when  its  resistance  is  computed  from 
the  mean  cross-section? 

Prob.  58-4.  A  110-volt  system  has  an  insulation  resistance 
for  each  wire  of  200  megohms  per  mile.  What  will  the  leak- 
age be  on  a  5-mile  line? 


THE  COMPUTATION  OF  RESISTANCE  105 

Prob.  69-4.  Insulation  resistance  should  be  high  enough 
so  that  not  more  than  one-millionth  of  the  rated  current  leaks 
through  the  insulation.  On  this  basis,  what  should  be  the 
insulation  resistance  per  mile  of  a  2-mile  line,  tramsmitting 
120  kilowatts  at  550  volts? 

Prob.  60-4.  The  resistance  of  a  car  heater  when  cold  (20°  C.) 
is  120  ohms.  If  the  temperature  rises  to  150°  C.  when  hot, 
how  much  less  current  does  it  take  when  hot  than  when  cold? 
The  material  of  the  heater  is  iron  wire.  The  voltage  is  550. 

Prob.  61-4.  A  rough  rule  for  the  safe  carrying  capacity  of 
copper  is  "  1000  amperes  per  square  inch  cross-section."  Ac- 
cording to  this  rule  what  should  be  the  diameter  of  a  round 
wire  capable  of  carrying  250  amperes? 

Prob.  62-4.  According  to  the  rule  in  Prob.  61-4,  what  should 
be  safe  carrying  capacity  of  No.  0000  (B.  &  S.)?  Compare  tho 
value  with  that  in  Table  III. 

Prob.  63-4.  Derive  the  equation  for  the  resistance  of  a 
round  rod  having  a  uniform  taper. 

Prob.  64-4.  It  is  desired  to  transmit  100  kilowatts  of  elec- 
trical power  to  a  small  town  and  to  deliver  it  at  220  volts. 
The  distance  is  4  miles  and  it  is  thought  that  the  line  will 
pay  if  an  amount  of  power  equal  to  10  %  of  that  delivered  is  al- 
lowed for  line  loss,  (a)  Find  the  size  wire  necessary  and  the  volt- 
age at  which  the  power  must  be  generated.  (6)  What  size  wire 
would  be  necessary  if  a  three-wire  (220-440  volts)  system  were 
used?  (c)  Compute  the  weight  of  copper  saved.  (Use  the  near- 
est sizes  of  wire,  B.  &  S.  gauge.)  (d)  Compare  the  copper  for 
these  two  cases  with  that  necessary  where  the  power  is  trans- 
mitted at  4400  volts  (alternating  current),  assuming  that  alter- 
nating-current power  equals  volts  times  amperes  and  that  the 
voltage  drop  in  the  alternating-current  line  is  all  resistance  drop. 

Prob  65-4.  A  transformer  is  an  electrical  appliance  con- 
sisting of  a  laminated  iron  core  around  which  are  wound  the 
primary  and  the  secondary  windings.  It  is  enclosed  in  a  pressed 
steel  tank  which  is  filled  with  oil.  The  heat  generated  in  the 
iron  core  and  copper  flows  into  the  oil  by  conduction  and  causes 
the  oil  to  circulate.  The  circulation  of  the  oil  causes  a  fairly 
uniform  distribution  of  temperature  throughout  the  transformer. 
Assume  that  the  temperature  is  the  same  at  all  points  within 
the  transformer  and  that  the  watts  radiated  per  unit  area  are 
proportional  to  the  temperature  difference.  The  results  of  a 
heat-run  on  a  certain  transformer  show  a  temperature  rise  of 


106      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

22°  C.  in  2  hours.  Heat  was  generated  within  the  transformer 
at  the  rate  of  225  joules  per  second.  The  radiating  surface  of 
the  transformer  is  10  square  feet.  The  heat  radiated  per 
square  foot  per  degree  of  temperature  difference  equals  0.5 
joule  per  second. 

(a)  With  a  room  temperature  of  25°  C.  what  will  be  the  tem- 
perature of  the  transformer  at  the  end  of  one  hour?  What 
will  be  the  ultimate  temperature? 

(6)  If  the  heat  is  generated  at  twice  the  above  rate,  what 
will  be  the  temperatures  asked  for  in  (a)? 

(c)  If  this  transformer  is  taken  out  of  circuit  when  its  tem- 
perature is  12°  C.  above  room  temperature,  how  long  will  it 
take  for  the  temperature  to  drop  6  degrees?     10  degrees? 

(d)  Construct    a   temperature-rise-versus-time    curve    for    a 
heat  input  of  200  watts. 

(e)  Construct    a    cooling    curve    (temperature-versus-time) 
starting  with  the  transformer  10°  C.  above  room  temperature. 

(/)  What  is  the  average  specific  heat  of  the  transformer 
material? 

Prob.  66-4.  A  Wheatstone  bridge  is  constructed  with  a  slide 
wire,  as  shown  in  Fig.  19.  Determine  the  relations  existing 
among  the  four  resistances  for  maximum  sensitivity  to  un- 
balance. 

Prob.  67-4.  One  of  the  cables  across  the  English  Channel  has  a 
conducting  core  of  #16  copper  wire  64  mils  in  diameter  (British  Stan- 
dard Gauge) .  It  is  fifty  miles  long,  and  may  be  assumed  to  follow 
a  path  which  is  approximately  parabolic,  reaching  a  maximum 
depth  of  one  mile.  The  temperature  of  the  water  varies 
linearly  with  the  depth,  and  may  be  taken  as  fifty  degrees  F. 
at  the  surface,  and  twenty-five  degrees  F.  at  the  max.  depth. 
Compute  the  total  resistance  of  the  cable. 


CHAPTER  V 
ELECTROLYTIC  CONDUCTION 

In  the  preceding  chapters  we  have  considered  conduction 
through  metals.  In  all  such  cases  the  wire  is  left  entirely 
unchanged  chemically  no  matter  how  long  the  current  is 
allowed  to  flow.  We  will  now  study  a  type  of  conduction 
in  which  the  flow  of  current  is  accompanied  by  a  chemical 
change  in  the  conductor. 

46.  Electrolytes  and  lonization.  Pure  water  is  a  fair 
insulator.  The  addition  of  a  minute  amount  of  a  soluble 
acid,  base  or  salt  will  produce  a  conducting  solution.  Cer- 
tain fused  salts  and  non-aqueous  solutions  are  also  conduct- 
ing. When  a  current  is  passed  through  a  liquid  of  this 
sort,  a  chemical  change  appears  in  the  neighborhood  of 
the  electrodes.  Such  liquids  are  called  electrolytes. 

Pure  water  is  an  insulator  because  there  are  no  carriers  of 
electricity  present.  When  a  salt  is  dissolved  in  the  water, 
there  still  are  no  free  electrons,  but  a  new  kind  of  carrier  of 
electricity  is  present.  This  is  the  ion.  When  a  quantity  of 
a  salt  is  dissolved,  part  of  the  molecules  separate  into  com- 
ponent parts.  These  are  not,  however,  the  atoms  of  which 
the  molecule  is  constructed.  The  parts  are  similar  to 
atoms  or  groups  of  atoms  except  that  so"me  have  extra 
attached  electrons  which  do  not  normally  belong  to  them 
and  others  have  too  few  electrons  to  be  normal.  These 
parts  are  called  ions  and  they  are  electrically  charged. 
An  ion  having  less  than  its  normal  number  of  electrons  is 
positively  charged  and  an  ion  having  more  than  its  normal 
number  is  negatively  charged.  When  an  electromotive 
force  is  applied  to  an  electrolyte,  the  ions  move  through 
the  liquid,  those  charged  positively  with  a  deficiency  of 

107 


108      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

electrons  moving  in  the  direction  of  the  applied  electromotive 
force,  and  the  negative  ions  charged  with  extra  electrons, 
in  the  opposite  direction.  An  ion  is  negatively  charged 
when  it  has  extra  electrons  attached  because  the  electrons 
themselves  are  negative  particles  of  electricity.  A  normal 
atom  is  made  up  of  a  so-called  nucleus  which  is  positively 
charged,  and  just  enough  negative  electrons  to  exactly 
balance  this  positive  charge.  Therefore  if  there  are  more 
than  this  number  of  (negative)  electrons  attached,  the  atom 
is  negatively  charged  and  becomes  a  negative  ion.  Similarly 
an  atom  lacking  in  (negative)  electrons  would  have  a  sur- 
plus of  positive  charge  and  would  become  a  positive  ion. 
Note  that  there  are  two  currents  flowing  through  the  fluid, 
one  made  up  of  ions  carrying  an  excess  of  small  particles  of 
electricity  (the  electrons),  the  other  made  up  of  ions  which 
are  deficient  in  these  particles  of  electricity.  The  total 
electric  current  flowing  is  the  sum  of  these  two.  It  will  be 
remembered  that  it  is  the  flow  of  the  "  excess  "  electrons 
only  which  constitutes  the  current  of  electricity  in  a  metallic 
conductor,  the  positive  charges  being  bound.  Thus,  curi- 
ously enough  and  rather  unfortunately,  this  flow  of  electrons 
is  in  the  direction  opposite  to  that  which  was  chosen  as  the 
positive  direction  before  the  electron  theory  was  known. 
Therefore  when  we  speak  of  the  flow  of  electricity  as  being 
in  a  certain  direction,  we  may  mean  that  there  is  a  current 
of  electrons  flowing  in  the  opposite  direction.  Of  course 
our  conventional  meanings  of  " positive"  and  " negative" 
make  a  current  of  negative  particles  in  one  direction  equiv- 
alent to  a  current  of  positive  particles  in  the  other  direction : 
thus  no  real  confusion  should  result. 

This  splitting  a  substance  into  positive  and  negative  ions 
involves  the  separation  of  the  parts  of  a  chemical  substance, 
and  chemical  effects  are  in  evidence  at  the  ends  of  the  elec- 
trolytic conducting  path.  These  chemical  effects  may  in- 
volve the  release  of  gas  bubbles  at  the  electrodes.  They 
may  produce  electroplating  or  cause  the  metal  of  the  elec- 


ELECTROLYTIC  CONDUCTION  109 

trodes  to  go  into  solution  or  bring  about  changes  in  the 
chemical  nature  of  the  electrolyte  near  the  electrodes. 

Electrolytic  conduction  is  made  use  of  in  primary  bat- 
teries and  in  storage  batteries,  for  the  plating  and  refining 
of  metals  and  for  the  production  of  various  chemicals. 
On  the  other  hand,  under  certain  conditions  it  will  produce 
harmful  effects  such  as  the  destruction  of  pipes  and  con- 
crete reinforcements  by  stray  currents  from  trolley  systems 
and  other  sources. 

47.  Electrolytes  and  Dissociation.  A  liquid  conductor 
in  which  conduction  takes  place  by  reason  of  the  presence 
of  charged  ions  is  called  an  electrolyte.  The  plate  or  electrode 
by  which  the  current  enters  the  solution  is  called  the  anode 
and  that  by  which  the  current  leaves  is  called  the  cathode. 
The  anode  is  hence  at  a  positive  potential  with  respect  to 
the  cathode.  Negatively  charged  ions  move  toward  the 
anode  and  positively  charged  ions  toward  the  cathode. 

When  an  ion  reaches  an  electrode  it  recovers  its  normal 
supply  of  electrons  and  becomes  an  atom.  As  a  free,  un- 
combined  atom  in  the  nascent  state,  it  usually  immediately 
combines  chemically  with  neighboring  atoms  or  molecules. 
This  combination  may  involve  atoms  of  its  own  kind,  those 
of  the  solution  or  those  of  the  electrodes.  Positive  ions 
arriving  at  the  cathode  take  electrons  from  it.  Negative  ions 
arriving  at  the  anode  give  up  electrons  to*  it.  The  electro- 
lyte as  a  whole  remains  uncharged  and  hence  just  as  many 
electrons  are  taken  from  the  cathode  as  are  passed  to  the 
anode.  The  electrons  which  pass  through  the  external 
metallic  circuit  are  hence  passed  through  the  solution  by 
being  carried  along  by  ions. 

Univalent  atoms  in  solution  become  singly  charged  ions. 

Thus  if  common  salt  is  dissociated  into  its  ions,  we  have 

NaCl^Na++  Cl~,*  (1) 

*  The  arrow  indicates  a  chemical  reaction.  When  two  arrows  are 
used,  they  indicate  a  reversible  reaction,  or  one  that  can  proceed  in 
either  direction  until  equilibrium  is  reached. 


110     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

which  means  that  a  molecule  of  sodium  chloride  splits  in- 
to a  positively  charged  ion  of  sodium  with  one  electron 
less  than  normal  and  a  negatively  charged  ion  of  chlorine 
with  one  excess  electron.  Polyvalent  atoms  give  rise  to 
ions  with  a  number  of  charges  equal  to  the  valency.  Thus 
if  calcium  chloride  is  dissolved  and  dissociated,  we  have 

CaCl2  +±  Ca  ++  +  Cl-  +  Cl~.  (2) 

One  molecule  of  the  salt  thus  produces  a  calcium  ion  with 
two  extra  positive  charges  and  two  singly  charged  negative 
chlorine  ions. 

Ions  are  not  necessarily  made  up  from  single  atoms. 
Chemical  groups  or  radicals  may  produce  single  so-called 
complex  ions.  Thus  when  ammonium  chloride  is  dis- 
sociated, we  have 

NH4C1  ?±  (NH4)+  +  C1-;  (3) 

that  is,  the  ammonium  radical  produces  a  single  positive 
ion.  On  the  final  dissociation  of  sulphuric  acid  we  obtain 

H2S04  +±  H+  +  H+  +  (S04)-  -;  (4) 

that  is,  two  positive  hydrogen  ions  and  a  doubly  charged 
negative  ion  formed  from  the  (SO4)  group. 

There  may  be  several  kinds  of  ions  in  a  solution  simul- 
taneously. Thus  if  sodium  chloride  and  potassium  bromide 
are  dissolved  together,  we  shall  have  positive  sodium  and 
potassium  ions  and  negative  chlorine  and  bromine  ions. 

Not  all  of  the  dissolved  substance  is  dissociated.  A 
greater  proportion  is  split  into  ions  when  the  solution  is 
dilute.  The  proportion  also  varies  greatly  with  the  sub- 
stance used.  Strong  acids  or  bases  are  most  highly  dis- 
sociated. An  acid  may  be  defined  as  a  substance  which 
produces  hydrogen  ions  in  solution;  a  base  as  a  substance 
which  produces  hydroxyl  ions,  that  is,  ions  of  the  form  (OH)~. 
A  strong  acid  such  as  nitric  acid  is  largely  dissociated, 
while  a  weak  acid  such  as  boric  acid  produces  relatively 
few  ions. 


ELECTROLYTIC  CONDUCTION  111 

The  conductivity  of  an  electrolyte  is  equal  to  the  current 
produced  by  unit  potential  gradient  (volts  per  centimeter) 
in  the  solution.  The  conductivity  is  the  sum  of  the  con- 
ductivities produced  by  each  sort  of  ion  separately.  These 
separate  conductivities  are  proportional  to  the  number  of 
ions  present  and  to  their  mobility,  that  is,  the  speed  with 
which  they  move  through  the  electrolyte  under  a  given 
impressed  voltage. 

A  strong  solution  of  hydrochloric  acid  is  highly  conduct- 
ing, while  a  strong  solution  of  boric  acid  is  of  comparatively 
high  resistance.  The  wide  difference  disappears  for  very 
dilute  solutions,  when  both  acids  are  largely  dissociated. 

A  normal  solution  of  hydrochloric  acid  has  a  resistivity 
of  about  3  ohm-centimeters  at  20°  C.  By  a  normal  solu- 
tion is  meant  one  in  which  the  number  of  grams  of  acid 
per  liter  of  solution  is  equal  to  the  molecular  weight  of  the 
acid  (in  this  case  36.5)  divided  by  the  number  of  replaced 
or  replaceable  hydrogen  atoms. 

Even  pure  water  is  not  a  complete  insulator.  Pure 
water  dissociates  very  slightly  into  hydrogen  and  nydroxyl 
ions. 

H20  =  H+  +  (OH)-.  (5) 

This  dissociation  is  so  slight  that  the  resistance  of  a  cubic 
centimeter  of  chemically  pure  water  is  approximately  0.5 
X  106  ohms  at  18°  C.* 

48.  Electric  Potential  Series.  If  the  electrodes  are  of 
different  metals,  the  voltage  between  them  will  not  be 
equal  simply  to  the  resistance  drop  in  the  electrolyte.  In 
fact,  if  two  plates  of  different  metals  are  dipped  in  an  elec- 
trolyte which  is  capable  of  combining  chemically  with  one 
of  them,  a  voltage  will  appear  between  the  plates,  even 
when  there  is  no  current  flowing.  This  chemically  pro- 
duced voltage  is  the  voltage  produced  and  used  in  primary 
and  in  storage  batteries. 

*  "  Handbook  of  Chemistry  and  Physics,"  Chemical  Rubber  Co. 


112      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

The  potential  produced  on  open  circuit  in  this  manner 
depends  simply  upon  the  metals  employed  and  not  upon 
the  electrolyte  so  long  as  the  latter  is  chemically  active 
with  respect  to  one  electrode. 

The  elements  may  be  arranged  in  a  table  called  the  po- 
tential series,  which  is  shown  below  for  some  of  the  more 
common  elements. 

TABLE 

POTENTIAL  SERIES  OF  ELEMENTS  COMPARED  WITH  HYDROGEN 

Manganese +  1 . 07  volts 

Zinc +  0.77 

Cadmium  +  0.42 

Iron +'0 . 35 

Cobalt +~0.23 

Nickel +  0.22 

Tin +  0.19 

Lead +  0.15 

Hydrogen 0 

Copper   -  0.33 

Mercury —  0. 75 

Silver -0.77 

If  two  substances  are  inserted  in  an  electrolyte,  the  chem- 
ical potential  between  them  may  be  found  from  the  above 
table  by  taking  the  algebraic  difference  between  the  voltages 
set  opposite  the  two  elements  in  question.  In  any  electro- 
lytic cell,  the  voltage  between  the  electrodes  will  then  be 
equal  to  this  voltage  plus  or  minus  the  resistance  drop  in 
the  electrolyte.  The  plus  sign  will  be  used  if  the  current 
is  caused  to  flow  in  a  direction  opposite  to  the  chemical 
voltage  and  the  minus  sign  if  the  current  is  in  the  reverse 
direction.  The  element  highest  up  in  the  table  will  be 
positive  and  hence  the  chemical  voltage  will  be  in  a  direc- 
tion toward  this  terminal. 

A  certain  care  must  be  used  in  employing  this  table  to 
make  sure  that  the  element  used  is  actually  the  one  forming 


ELECTROLYTIC  CONDUCTION  113 

the  surface  of  the  electrode.  Suppose  that  two  nickel 
electrodes  are  dipped  in  a  solution  of  zinc  sulphate  and  a 
current  is  passed  through  the  combination.  At  the  first 
instant  there  will  be  no  chemical  voltage,  for  the  electrodes 
are  both  of  nickel.  After  the  current  has  flowed  for  a  very 
short  time,  however,  this  will  no  longer  be  the  case.  In  any 
solution  of  zinc  sulphate  there  will  be  some  positive  ions 
(Zn)++  and  some  negative  ions  (SO4)~~. 

The  positive  zinc  ions  arriving  at  the  cathode  will  be  dis- 
charged and  become  atoms  of  zinc  which  will  adhere  to  the 
cathode.  The  cathode  will  soon  become  covered  with  a 
plated  layer  of  zinc  and  we  shall  have  the  zinc-nickel  chem- 
ical voltage  of  0.55  volt  opposing  the  flow  of  current. 

In  many  cases,  a  gas  such  as  hydrogen  will  collect  at  one 
of  the  electrodes.  This  gas  in  a  thin  layer  completely 
covers  the  electrode  below  the  surface  of  the  liquid.  We 
then  have  practically  a  hydrogen  electrode  as  far  as  the 
chemical  voltage  is  concerned. 

Prob.  1-6.  Describe  how  to  make  a  normal  solution  of  sul- 
phuric acid :  a  one-tenth  normal  solution. 

Prob.  2-5.  A  normal  solution  of  H2S04  has  a  conductivity  of 
0.198  mho-centimeter  at  18°  C.  A  tV  normal  solution  has  a  con- 
ductivity of  0.0225,  a  ifa  normal  solution  0.00308,  a  y^Vu"  normal 
solution  0.000361,  all  under  the  above  conditions.  Make 
a  table  and  curve  showing  the  relation  between  concentration 
(in  percentage  normal)  of  sulphuric  acid  and  the  "  equivalent 
conductivity  "  —  that  is,  the  conductivity  divided  by  the 
concentration. 

Prob.  3-5.  From  the  curve  of  Prob.  2-5,  what  statement 
can  be  made  in  regard  to  the  dissociation  of  sulphuric  acid 
in  aqueous  solution? 

Prob.  4-5.  By  using  the  curve  of  Prob.  2-5,  find  the  con- 
ductivity of  a  solution  formed  by  adding  150  grams  of  sul- 
phuric acid  to  one  liter  of  water. 

Prob.  5-5.  If  the  solution  of  Prob.  4-5  were  used  between 
two  storage-battery  plates  6  inches  square  and  separated 


114      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

|  inch,  what  would  be  the  resistance  of  the  electrolyte  between 
the  plates? 

49.  Quantity  Relations.  Faraday  discovered  that  when 
a  current  of  electricity  is  passed  through  an  electrolyte,  the 
amount  of  chemical  action  produced  is  proportional  to  the 
quantity  of  electricity  passed.  That  is,  for  example,  if  we 
are  plating  nickel,  the  amount  of  nickel  deposited  is  pro- 
portional to  Q,  where 

Q  =  IT,  (6) 

provided  an  unvarying  current  I  flows  for  a  time  T.  If 
the  current  i  is  not  constant,  the  quantity  of  nickel  deposited 
is  proportional  to  Q,  where 

Q  =    Fid*.  (7) 

Jo 

This  fact  is  made  use  of  in  the  U.  S.  legal  definition  of  an 
ampere  current  of  electricity,  see  p.  19. 

He  also  discovered  that  the  weight  of  metal  deposited 
or  of  gas  liberated  at  an  electrode  by  a  given  amount  of 
electricity  is  proportional  to  the  atomic  weight  of  the  substance 
deposited  or  liberated,  provided  the  valence  is  the  same  in 
each  case.  The  valence  is  the  number  of  hydrogen  atoms 
with  which  one  atom  of  the  substance  in  question  forms  a 
stable  compound. 

Thus  an  ampere  flowing  for  one  hour  in  a  nickel  plating 
bath  will  deposit  1.09  grams  of  nickel,  while  the  same  cur- 
rent in  the  same  length  of  time  will  deposit  1.22  grams  of 
zinc,  the  solution  being  of  a  bivalent  salt  in  each  case. 
The  ratio  1.09  :  1.22  is  the  same  as  the  ratio  58.7  : 65.4  of 
the  atomic  weights  of  the  two  metals  considered.  If  the 
valence  is  different  for  the  ions,  the  amount  deposited  is 
proportional  to  the  atomic  weight  divided  by  the  valence. 

These  laws  may  all  be  summed  up  in  the  single  expression 

wQ 
= 


ELECTROLYTIC  CONDUCTION  115 

where 

m  is  the  mass  in  grams  deposited  or  liberated, 

Q  is  the  quantity  of  electricity  in  coulombs  passed 

through  the  solution, 

w  is  the  atomic  weight  of  the  element  under  consider- 
ation, 
n  is  the  valence  of  the  element. 

The  constant 

"  (8) 

96,540  n 

is  called  the  electrochemical  equivalent  of  the  given  element. 
It  is  the  mass  of  the  element  in  grams  deposited  by  one 
3oulomb. 

The  constant  96,540  was  determined  experimentally. 
It  is  the  number  of  coulombs  necessary  to  deposit  or  liberate 
an  amount  of  a  univalent  element  in  grams  numerically 
equal  to  its  atomic  weight,  that  is,  a  "gram -atom"  of 
the  substance.  Thus,  for  instance,  it  is  the  amount  of 
electricity  necessary  to  liberate  one  gram  of  hydrogen  at 
the  cathode  in  decomposing  water.  The  same  amount  of 
electricity  will,  of  course,  liberate  \6  or  8  grams  of  oxygen 
at  the  anode,  since  the  atomic  weight  of  oxygen  is  16  and 
the  valency  is  2.  This  large  quantity  of  electricity  is 
called  a  "  faraday." 

1  faraday  =  96,540  coulombs. 
It  will  be  remembered  that  there  are  approximately 

6.3  X  1018 

electrons  in  a  coulomb.  Multiplying,  we  see  that  one 
faraday  corresponds  to 

6.1  X  1023  electrons 

moved  along  the  circuit.  If  a  univalent  element  is  being 
dealt  with,  each  electron  moved  through  the  circuit  cor- 


116      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

responds  to  one  ion  discharged  and  become  an  ordinary 
atom  which  may  then  enter  into  some  chemical  combination 
or  become  an  atom  of  gas.  The  very  large  number  of 
electrons  given  above  is  hence  also  equal  to  the  number  of 
atoms  in  a  gram-molecule  of  a  substance;  for  example,  the 
number  of  hydrogen  atoms  in  a  gram  of  hydrogen  gas. 
This  number  is  therefore  a  very  important  physical  con- 
stant. 

The  above  rules  may  be  applied  also  to  combinations  of 
atoms  which  appear  as  ions.  Thus  when  sulphuric  acid 
is  dissolved  in  water,  we  have  (S04)  ~  "  ions.  Their  valency 
is  2,  since  each  one  combines  with  two  hydrogen  atoms. 
The  molecular  weight  is  equal  to  that  of  sulphur  plus  four 
times  that  of  oxygen,  or 

32  +  (4  X  16)  =  96. 
The  electrochemical  equivalent  is 

96 
2  X  96,540 

or  approximately  0.0005.  One  coulomb  of  electricity  hence 
discharges  0.0005  gram  of  (S04)  ions  at  the  anode  when 
passed  through  such  a  solution.  These  discharged  ions 
enter  immediately  into  chemical  combination,  for  S04  can- 
not exist  uncombined. 

In  applying  the  above  formula  for  quantity  relations, 
great  care  must  be  used  in  several  ways  if  correct  results 
are  to  be  obtained.  Auxiliary  reactions  often  take  place 
and  bring  the  results  into  error.  The  auxiliary  reactions 
may  be  due  to  several  causes.  Several  ions  may  be  pass- 
ing through  the  solution  toward  a  single  one  of  the  elec- 
trodes at  the  same  time.  In  this  case  the  total  number  of 
ions  passed  will  be  governed  by  the  above  rule,  but  the  rule 
does  not  indicate  what  will  be  the  proportion  of  each  kind. 
This  is  determined  by  the  mobilities  of  the  different  ions. 
The  rule  must  therefore  be  used  with  great  care,  and  only 


ELECTROLYTIC  CONDUCTION  117 

where  there   is   a   thorough   knowledge   of  the   chemistry 
involved. 

Prob.  6-5.  Zinc  and  copper  strips  are  dipped  in  a  solution 
of  sulphuric  acid  to  form  a  simple  primary  cell.  What  voltage 
will  it  give  at  first? 

Prob.  7-5.  After  the  current  has  flowed  for  a  very  short 
time  in  the  above  cell,  one  of  the  electrodes  will  become  covered 
with  a  layer  of  hydrogen  which  will  prevent  the  electrolyte  from 
touching  the  metal  underneath.  What  will  then  be  the  voltage 
given  by  this  polarized  cell? 

Prob.  8-5.  In  a  plant  for  refining  copper  an  electric  current 
is  passed  through  a  solution  of  copper  sulphate  for  twenty- 
four  hours.  The  amount  of  copper  deposited  in  this  time  is 
2.65  pounds.  What  average  current  was  used? 

Prob.  9-5.  Two  electro-plating  vats  are  arranged  in  series; 
one  is  for  nickel  plating  and  the  other  for  silver  plating.  If 
3.26  ounces  of  silver  are  deposited  in  a  given  time,  how  much 
nickel  is  deposited  in  the  same  time? 

Prob.  10-5.  How  many  electrons  per  second  are  moved 
when  one  ampere  current  flows? 

50.  Primary  Cells.  If  a  strip  of  copper  and  a  strip  of 
zinc  are  dipped  into  a  weak  solution  of  sulphuric  acid,  we 
have  a  battery.  The  chemical  voltage  of  the  Cu-Zn  com- 
bination of  1.10  volts,  as  taken  from  the  table  of  Article 
48,  will  be  found  to  exist  between  the  electrodes.  The 
copper  will  be  positive.  If  the  two  electrodes  are  con- 
nected by  an  external  circuit,  a  current  will  flow  in  accord- 
ance with  Ohm's  law.  The  internal  resistance  of  the  cell 
will  depend  upon  the  size  and  position  of  the  electrodes  and 
the  concentration  of  the  electrolyte. 

There  are  (H)+  and  (S04)~  ~  ions  in  the  solution.  When 
current  flows,  hydrogen  will  appear  at  the  copper  cathode 
in  bubbles  of  gas  in  an  amount  determined  by  Faraday's 
laws.  The  (S04)~  ~  IODS  will  arrive  at  the  zinc  anode,  be- 
come discharged  and  immediately  combine  with  atoms  of  zinc 
to  form  zinc  sulphate,  ZnSO4,  which  will  instantly  dissolve 
in  the  solution.  The  amount  of  zinc  dissolved  will  hence 
also  be  found  by  Faraday's  law. 


118      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

There  are,  however,  several  difficulties  with  this  sort  of 
cell.  First,  it  polarizes.  That  is,  after  the  current  has 
been  allowed  to  flow  a  short  time,  the  copper  becomes  cov- 
ered with  a  layer  of  hydrogen,  and  we  have  only  the  chemi- 
cal voltage  of  an  H-Zn  combination,  about  0.77  volt.  A 
greater  disadvantage  is  the  fact  that  the  layer  of  gas  in- 
creases the  internal' resistance  of  the  cell  until  it  becomes 
very  high. 

Second,  the  zinc  will  be  dissolved  by  what  is  called  local 
action,  which  takes  place  in  the  following  manner.  Differ- 
ent parts  of  the  zinc  are  of  different  purity.  These  form 
small  local  short-circuited  electrolytic  cells,  which  cause  the 
zinc  to  be  rapidly  attacked.  This  is,  in  fact,  the  way  in 
which  metals  are  ordinarily  corroded. 

Third,  the  liquid  is  easily  spilled. 

The  usual  form  of  commercial  primary  battery  is  the  so- 
called  "  dry  cell."  Polarization  is  prevented  by  including 


FIG.  46.    The  cross-seotion  of  a  dry  cell. 

certain  chemicals,  such  as  manganese  dioxide,  which  com- 
bine with  or  absorb  the  gases  as  fast  as  they  are  formed. 
Carbon  and  zinc  electrodes  are  used,  the  zinc  forming  the 
container  for  the  electrolyte.  Local  action  is  prevented 
by  using  certain  zinc  alloys  which  are  very  homogeneous. 
The  electrolyte,  usually  ammonium  chloride,  is  absorbed 
in  blotting  paper  or  plaster  paris,  and  the  whole  is  hermet- 
ically sealed  with  a  tar  compound  to  prevent  spilling.  A 
section  of  such  a  cell  is  shown  in  Fig.  46. 

Such  cells  are  convenient  where  small  amounts  of 
power  are  required  in  easily  available  form,  as  for  door  bells, 
flash  lights,  gas-engine  ignition  and  so  on.  When  any  large 


ELECTROLYTIC  CONDUCTION  119 

quantity  of  electric  energy  is  needed,  it  is  too  expensive  to 
obtain  it  in  this  manner.  A  short  computation  of  the  cost 
of  zinc  alone  will  show  this. 

Assume  a  dry  cell  to  give  1.5  volts  terminal  voltage. 
To  obtain  one  kilowatt-hour  from  dry  cells  of  this  sort 
connected  in  parallel  would  require 

1000  X  3600       0  . 
-  —=—    -  =  2.4  X  106  coulombs. 
l.o 

Such  a  quantity  will,  in  accordance  with  Faraday's  laws, 
require  the  consumption  of  a  quantity  of  zinc  in  grams 
equal  to 

2.4  X  106  X  65.4 


96540X2 
or 

1.8  pounds. 

At  30^  a  pound  for  zinc,  the  energy  thus  obtained  would 
cost  54^  per  kilowatt-hour  for  the  zinc  consumed  alone. 
Other  costs,  such  as  that  of  fabrication,  would  be  even 
larger. 

The  electrical  energy  obtained  comes  from  the  conversion 
of  the  chemical  energy  released  in  combining  the  zinc  with 
sulphuric  acid  to  form  zinc  sulphate.  Energy  obtained  in 
this  way  from  a  metal  is  thus  inherently  expensive,  for 
it  costs  more  to  form  sulphates  from  metals  than  to  form 
oxides  from  coal. 

Prob.  11-6.  The  zinc  plate  of  a  certain  battery  cell  weighs 
3.52  ounces.  How  many  hours  will  the  battery  cell  last  if  it 
is  required  to  deliver  an  average  current  of  2.14  amperes? 

Prob.  12-5.  A  cell  is  desired  which  will  deliver  an  average 
of  0.08  ampere  for  four  months.  What  weight  should  the  zinc 
plates  have? 

Prob.  13-5.  (a)  How  many  faradays  of  electricity  are 
moved  in  the  cell  of  Prob.  12-5?  (6)  How  many  electrons? 

51.  Storage  Batteries.  A  storage  cell  is  similar  in  action 
to  a  primary  cell,  except  that  the  chemical  actions  involved 


120      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

are  completely  reversible.  After  the  cell  is  discharged, 
the  passage  of  current  in  the  reverse  direction  restores  the 
electrodes  and  the  electrolyte  to  their  original  condi- 
tions. 

If  there  were'no  secondary  chemical  actions,  the  same  num- 
ber of  ampere-hours  would  need  to  be  passed  through  on 
charge  as  were  taken  out  on  discharge.  Practically,  a 
somewhat  greater  quantity  of  charge  is  necessary.  An 
ampere-hour  is  3600  ampere-seconds  or  3600  coulombs. 
Storage  batteries  are  usually  rated  in  ampere-hours  to  give 
the  quantity  of  electricity  which  can  be  taken  from  them 
without  discharging  them  below  a  safe  condition. 

There  are  two  commercial  forms  of  storage  batteries,  the 
lead  battery  and  the  nickel-iron  battery.  Each  has  its 
particular  field  of  usefulness.  The  lead  battery  is  capable 
of  passing  heavier  currents  for  a  short  period  and  is  hence 
used  for  automobile  starting.  It  is  of  lower  internal  resist- 
ance. At  the  present  time  the  lead  cell  is  cheaper  in  first 
cost  and  is  extensively  used  for  stationary  power  batteries, 
in  direct-current  power  and  lighting  service,  for  absorbing 
the  peak  load  and  for  short  stand-by  service  in  case  of 
break-down  of  generators.  The  principal  nickel-iron  or 
alkaline  battery  is  the  Edison  battery.  Its  uses  have  not 
as  yet  been  completely  developed.  On  account  of  lightness 
and  the  ability  to  withstand  shocks,  it  is  extensively  used  as 
a  vehicle  battery. 

The  lead  battery  illustrated  in  Fig.  47  has  plates  of 
lead  peroxide,  positive,  and  metallic  lead,  negative.  The 
electrolyte  is  dilute  sulphuric  acid.  It  will  be  noted  that 
neither  of  the  plates  is  soluble  in  the  electrolyte.  The 
lead  peroxide,  Pb(>2,  is  moulded  into  a  grid  of  lead  in 
the  "  pasted  "  type  of  cell,  and  is  formed  in  place  chem- 
ically in  the  "Plante"  type.  The  negative  plate  is  com- 
posed of  lead  in  a  spongy  condition  so  that  it  offers  a 
very  large  surface  to  the  electrolyte. 

On  discharging  the  battery  the  materials  of  both  plates  are 


ELECTROLYTIC  CONDUCTION  121 

converted  to  lead  sulphate.  The  simplest  chemical  reaction 
is 

Pb02  +  Pb  +  2  H2SO4  *±  2  PbSO4  +  2  H2O.         (9) 

This  may  be  read  either  way,  to  the  right  for  discharge  and 
to  the  left  for  charge.  As  a  matter  of  fact,  this  simple 
equation  does  not  completely  express  all  that  takes  place, 


FIG.  47.    Cut-away  view  showing  the  construction  of  a  lead 
storage  cell.     Electric  Storage  Battery  Co. 

and  the  chemistry  of  the  cell  is  quite  complicated,  especially 
in  the  presence  of  impurities. 

The  water  formed  on  discharge  further  dilutes  the  elec- 
trolyte. For  this  reason  the  specific  gravity  of  the  electro- 
lyte as  measured  by  a  hydrometer  is  an  accurate  measure  of 
its  condition,  provided  the  concentration  is  properly  ad- 
justed when  the  battery  is  fully  charged. 

The  voltage  delivered  by  a  lead  battery  varies  with  its 
charge.  When  fully  charged  each  cell  will  deliver  about 
2.1  volts  on  normal  load.  When  this  has  dropped  to  about 
1.8  volts,  the  battery  is  considered  fully  discharged.  Over- 
discharge  beyond  this  point  is  likely  to  form  sulphate  in 
such  form  that  it  cannot  be  converted  back  to  active  ma- 


122      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


terial  on  charge.  This  is  called  "  sulphating  "  of  the  bat- 
tery. On  charge  the  voltage  is  higher  than  on  discharge, 
partly  on  account  of  RI  drop  in  the  electrolyte  and  partly 
on  account  of  secondary  chemical  reactions.  It  rises  from 
about  2.1  to  2.6  volts  per  cell.  Typical  charge  and  dis- 
charge curves  are  shown  in  Fig.  48  for  normal  current  in 
each  case.  Since  the  current  is  constant,  the  efficiency  of 


2.60 
2.50 
2.40 
2.30 
2.20 

Ikio 

> 

2.00 

1.90 
1.80 
1.70 
1.60 


4         5 
Hours 


FIG.  48.  Curves  for  a  lead  cell  showing  the  relation  between 
terminal  voltage  at  normal  current  and  time  of  charge  and 
discharge. 

the  battery  may  be  found  by  dividing  the  area  under  the 
discharge  curve  by  the  area  under  the  charge  curve. 

Large  plates  placed  very  close  together  are  used  to  keep 
down  the  internal  resistance,  which  is  remarkably  low. 
On  short  circuit  several  hundred  amperes  may  be  drawn 
from  a  single  cell  of  forty-ampere-hour  size.  An  excessive 
current  of  this  sort  will,  however,  be  likely  to  cause  the 
plates  to  buckle  or  to  sulphate  on  the  surface.  This  last 


ELECTROLYTIC  CONDUCTION 


123 


effect  is  due  to  the  fact  that  on  heavy  currents  the  elec- 
trolyte in  the  interior  of  the  plates  becomes  very  dilute, 
and  does  not  have  time  to  become  the  same  by  diffusion 
as  the  main  body  of  the  liquid. 


POSITIVE  POLE 


SPENS.ON  »OM 


FIG.  49.     Cut-away  view  of  an  Edison  storage  cell.     Edison  Storage 

Battery  Co. 

The  weakness  of  lead  structurally  and  its  great  weight 
gave  considerable  difficulty  in  the  construction  of  plates 
that  would  not  sag  or  buckle,  and  that  would  retain  the 
active  material  in  good  electrical  contact.  Great  ingenuity 
has  been  shown  in  the  form  of  lead  grid  used  to  make  plates. 

The  alkaline  cell  is  principally  the  outcome  of  attempts  to 
overcome  the  weight  and  mechanical  weakness  of  the  lead 
plate.  The  Edison  cell  illustrated  in  Fig.  49  uses  elec- 


124      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

trodes  of  nickel  peroxide  for  the  positive  plate  and  finely 
divided  iron  for  the  negative.  The  electrolyte  is  a  21% 
solution  of  potassium  hydroxide,  with  some  lithium  hydrate. 
The  iron  has  a  small  amount  of  mercury  added  to  it  to 
render  it  more  readily  active,  and  there  are  flakes  of 
nickel  added  to  the  nickel  peroxide  to  increase  the  conduc- 
tivity. 

The  chemistry  of  this  cell  is  very  complicated  and  not 
entirely  understood.  The  products  of  discharge  are  iron 
oxide  and  a  further  oxide  of  nickel.  The  principal  reaction 
is 

5Fe+5Ni203-6H20+9H20^10Ni(OH)2H-5Fe(OH)2.  (10) 

Read  from  left  to  right  for  discharge  and  from  right  to  left 
for  charge.  The  electrolyte  is  not  changed  in  composition 
or  in  concentration  by  the  action  and  hence  the  specific 
gravity  is  not  an  indication  of  the  state  of  charge.  The 
way  in  which  the  electrolyte  enters  into  the  action  without 
becoming  changed  is  as  follows,  expressed  in  simple  form. 
The  ions  in  the  solution  are  K+  and  (OH)~.  The  positive 
plate  is  that  by  which  the  current  enters  on  charge  and 
leaves  on  discharge.  On  discharge  the  K  ions  move  toward 
the  positive  plate.  Then 

4  K  +  3  NiO2  -»  Ni304  +  2  K2O.  (11) 

At  the  other  plate  the  (OH)  ions  on  arriving  give 

8  (OH)  +  3  Fe  ->  Fe304  +  4  H2O.  (12) 

The  K20  is  unstable  and  immediately  after  forming,  decom- 
poses water  of  the  solution,  thus: 

4  K20  +  4  H20  -»  8  KOH.  (13) 

Thus  the  electrolyte  is  left  unchanged.  The  charge  and 
discharge  curves  and  the  voltages  obtained  are  shown  in 
Fig.  50. 

One  intermediate  reaction  is  needed  to  explain  several 
phenomena  peculiar  to  this  cell. 


ELECTROLYTIC  CONDUCTION 


125 


On  charge,  Ni02  is  first  formed  which  gradually  decomposes 
into  the  lower  oxide  Ni2O3  according  to  the  equation 


4  Ni02  =  2  Ni203  +  O2 


(14) 


giving  off  the  oxygen.  This  decomposition  accounts  for  the 
continual  gassing  which  takes  place  during  charge  and  even 
continues  for  considerable  time  after  charging  is  discontinued. 


2.00 


0.80 


1  2  3  4  5 

Hours  Charge  or  Discharge  at  Normal  Rate 


FIG.  50.     Curves  for  an  Edison  cell  showing  the  relation  between  ter- 
minal voltage  at  normal  current  and  time  of  charge  and  discharge. 


If  a  cell  is  allowed  to  stand  idle  after  a  charge,  it  will  take 
one  or  two  days  for  the  higher  oxide  to  completely  decompose 
into  Ni203.  This  reduction  is  much  hastened  by  discharge, 
since  any  Ni02  existing  in  the  positive  plate  will  be  used  up  first. 
This  rapid  reduction  of  Ni02  in  discharge  probably  accounts 
for  the  marked  drop  in  voltage  which  takes  place  at  the  be- 
ginning of  the  discharge  of  a  freshly  charged  cell.  It  also 
accounts  for  the  higher  capacity  obtained  when  a  cell  is  dis- 
charged immediately  after  being  charged. 

Prob.  14-6.  A  lead  storage  cell  has  an  internal  resistance  of 
0.002  ohm.  What  will  be  the  difference  in  its  terminal  voltage 
in  charge  and  discharge  at  40  amperes? 

Prob.  16-5.  A  280-ampere-hour  lead  storage  vehicle  bat- 
tery has  an  average  voltage  of  32  volts.  How  much  energy 
can  be  drawn  from  it? 

Prob.  16-6.  If  the  average  voltage  required  during  charge 
of  the  battery  of  Prob.  15-5  is  40  volts  and  the  time  of  charge 
10%  longer  than  the  time  of  discharge  at  normal  rate,  what  is 
the  efficiency  of  the  battery? 


126      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  17-5.  The  vehicle  of  Prob.  15-5  requires  six  horse 
power  output  to  drive  it  at  20  miles  per  hour  and  the  motors 
are  60%  efficient,  (a)  How  far  can  it  be  driven  at  this  speed 
on  one  charge?  (6)  What  is  the  cost  of  energy  per  mile  at  3^  per 
kilowatt-hour? 

52.  Electrolytic  Refining  of  Metals.  If  both  electrodes 
are  of  the  same  metal,  there  is  very  little  chemical  voltage 
present.  The  only  electrical  energy  used  in  passing  a 
current  through  the  combination  is  that  lost  in  heating  the 
electrolyte.  There  will  always  be  a  minute  chemical  volt- 
age due  to  differences  in  purity  of  the  metals  and  to  differ- 
ences in  concentration  in  the  parts  of  the  solution. 

In  the  electro-deposition  of  metals  for  plating  articles, 
the  cathode  is  always  the  article  to  be  plated,  the  electro- 
lyte a  dissolved  salt  of  the  metal  to  be  plated  and  the  anode  a 
piece  of  the  same  metal  or  an  inert  conductor.  In  the  latter 
case  there  is  a  chemical  voltage  and  the  electrolyte  must 
be  renewed  from  time  to  time.  There  are  many  "  kinks  " 
to  good  plating  technique.  Voltages  from  one  to  six  are 
usually  used. 

Copper,  nickel  and  other  metals  are  refined  by  plating 
them.  Electrolytic  copper  thus  prepared  is  very  pure  and 
is  used  for  wires,  sheets  and  many  other  purposes  where 
even  small  impurities  are  undesirable. 

In  copper  refining,  the  electrolyte  is  copper  sulphate. 
The  impure  metal  is  used  as  the  anode  and  a  thin  sheet  of 
pure  copper  is  used  as  a  starter  at  the  cathode.  A  small 
amount  of  common  salt  is  used  to  precipitate  any  silver 
which  dissolves  from  the  anode  and  the  silver  thus  recovered 
pays  a  large  part  of  the  costs  of  the  process.  Over  a  million 
tons  of  copper  are  annually  refined  in  this  manner.  The 
impurities  are  insoluble  in  the  solution  or  are  made  so  by 
the  addition  of  certain  chemicals,  and  thus  do  not  plate  out 
but  appear  as  a  mud  in  the  bottom  of  the  tank.  The  copper 
obtained  is  about  99.95%  pure,  the  chief  impurity  being 
hydrogen  dissolved  in  the  metal. 


ELECTROLYTIC  CONDUCTION  127 

The  electrolyte  is  used  hot  in  large  cells  which  are  usually 
connected  in  series  to  give  a  convenient  voltage  to  handle 
and  generate.  A  current  density  of  about  20  amperes  per 
square  foot  of  cathode  is  used.  From  Faraday's  laws, 
about  0.0026  pound  of  copper  is  deposited  per  ampere- 
hour.  Thus  to  deposit  a  ton  of  copper  requires  770,000 
ampere-hours,  or  for  T^  ton  per  cell  per  day,  about  3200 
amperes.  If  this  were  delivered  at  100  volts  for  a  long  series 
of  cells,  a  320-kilowatt  generator  would  be  necessary.  This 
would  supply  about  50  cells,  giving  5  tons  of  metal  per  day. 
Several  series  may  be  used  on  one  large  generator.  If  the 
above  electrical  energy  is  estimated  at  two  cents  per  kilo- 
watt-hour, the  cost  per  pound  for  energy  alone  is  0.15  cents 
per  pound.  A  considerable  fraction  of  the  cost  of  electro- 
lytic copper  refining,  strange  to  say,  is  the  interest  on  the 
investment  in  copper  tied  up  in  the  process. 

Prob.  18-5.  How  long  must  a  current  of  500  amperes  run 
in  an  electro-refining  vat  to  deposit  enough  copper  to  make  one 
mile  of  No.  00  wire? 

Prob.  19-6.  Zinc-coated  iron  is  commercially  known  as 
"  galvanized  iron."  How  thick  a  plate  of  zinc  will  be  put  on  a 
sheet  of  iron  having  40  square  feet  (both  sides)  if  a  current  of 
12  amperes  runs  for  16  hours  through  the  zinc-plating  solution? 

Prob.  20-5.  If  the  electric  energy  of  Prob.  18-5  costs  1.25^ 
per  kilowatt-hour  and  the  refined  copper  is  worth  I8i  per  pound, 
at  what  voltage  must  the  refining  vat  be  run  to  make  the  cost 
of  energy  equal  5%  the  cost  of  copper  refined? 

53.  Electrolysis.  Electrolytic  processes  are  useful  in 
many  ways.  They  enter  into  the  experience  of  an  electrical 
engineer  in  one  way  which  is  not  desirable.  This  is  in 
electrolysis. 

The  earth  is  a  conductor.  Moist  earth  with  some  dis- 
solved salts  present  acts  as  an  electrolyte.  The  resistivity 
of  this  electrolyte  is  high,  but  the  earth  is  so  large 
that  the  total  resistance  of  the  earth  between  two  points 
may  be  low.  In  fact,  the  resistance  between  two 


128      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

plates  buried  in  the  earth  will  be  found  to  be  due  almost 
entirely  to  the  current  paths  near  the  electrodes  and  prac- 
tically independent  of  the  distance  between  the  plates. 
The  resistance  of  the  ground  return  of  a  telegraph  circuit 
will  be  in  the  neighborhood  of  twenty  ohms,  with  well 
constructed  "  grounds  "  or  connections  to  the  earth. 

Where  a  trolley  system  is  operated  with  a  rail  return  for 
the  current,  large  stray  currents  in  the  earth  will  always 
be  found.  These  do  no  harm  except  where  they  take  short 
cuts  through  sections  of  metal.  They  do  no  harm  here 
where  they  enter  the  metal,  but  where  the  current  leaves, 
the  metal  is  dissolved  or  electrolysed.  Water  pipes  are 
particularly  subject  to  difficulty  in  this  regard  and  will 
soon  be  eaten  through  and  leak  if  large  currents  pass.  Great 
care  must  be  used  also  that  there  is  not  electrolysis  of  the 
reinforcing  rods  of  concrete  structures,  for  the  failure  of 
the  structure  may  follow.  The  electrolysis  of  metal  pipes 
buried  in  the  earth  has  been  the  cause  of  much  litigation. 

Stray  currents  from  trolley  systems  may  be  largely 
minimized  by  the  use  of  negative  feeders,  that  is,  of  large 
conductors  connecting  to  the  rail  system  to  carry  the  current 
from  points  out  on  the  line  back  to  the  power  house.  These 
negative  feeders  may  also  be  connected  to  pipes  to  protect 
them,  but  care  must  be  taken  here  that  more  harm  is  not 
done  than  good,  due  to  the  resistance  of  leaded  or  gasket 
joints  of  high  resistance  in  the  pipe  line. 

The  amount  of  metal  dissolved  by  electrolysis  with  a 
given  current  may  be  computed  roughly  by  Faraday's  laws. 
Secondary  reactions  prevent  great  accuracy. 

There  is  electrolysis  even  when  alternating  current  flows, 
but  it  is  usually  less  than  1%  of  what  would  be  caused  by 
the  same  amount  of  direct  current.  With  alternating  cur- 
rent the  metal  is  dissolved  on  one  half  cycle  and  plated 
back  on  the  next.  Not  all  that  is  dissolved  will  plate  back, 
even  when  the  two  half  cycles  of  current  are  identical,  due 
to  diffusion  of  the  salt  formed  and  to  secondary  reactions. 


ELECTROLYTIC  CONDUCTION 


129 


With  alternating  current  the  difficulty  will  occur  both 
where  the  current  leaves  and  where  it  enters  the  pipe,  for 
these  conditions  are  reversed  when  the  direction  of  the 
current  changes. 


FIG.  51.  Nine  2500-kw.  converters  used  to  supply  current  for  the 
production  of  aluminum.  There  are  thirty-six  similar  machines 
installed  in  the  Marysville,  Tenn.  Plant  of  the  Aluminum  Co.  of 
America.  Westinghouse  Electric  and  Mfg.  Co. 

54.  Other  Electrochemical  Processes.  Electrochemistry 
is  a  profession  by  itself.  It  is  rapidly  becoming  of  funda- 
mental importance  in  our  complex  life.  A  large  number  of 
chemical  processes  are  now  conducted  by  the  aid  of  elec- 
tricity and  can  only  be  given  passing  mention  here. 

Fused  salts  in  place  of  salts  dissolved  in  water  may  be 
decomposed  by  electrolytic  action.  This  often  requires 
the  high  temperature  obtained  from  the  electric  furnace. 
In  other  cases,  moderate  temperatures  may  be  used. 

Aluminum  is  manufactured  in  this  manner  by  electrically 
decomposing  an  ore  of  aluminum  such  as  bauxite, 


130       PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Aluminum  requires  the  largest  amount  of  electricity  per 
pound  of  any  metal  obtained  commercially  by  electrolysis. 
Fig.  51  shows  part  of  a  battery  of  thirty-six  2500-kilowatt 
converters  used  to  supply  current  to  one  plant  producing 
aluminum.  The  cost  of  aluminum  is  largely  determined  by 
the  cost  of  electric  energy.  Metallic  sodium  and  potassium 
are  similarly  produced,  but  by  the  use  of  much  higher 
temperature. 


FIG.  52.  Cell  room  showing  six  banks  of  74  cells  each.  United  States 
Government  Chlorine-Caustic  Soda  Plant,  Edgewood  Arsenal,  Md. 
Hon.  Benedict  Crowell  in  "America's  Munitions." 

Gaseous  hydrogen  and  oxygen  are  commercially  produced 
in  one  process  by  the  electrolytic  decomposition  of  water. 

The  fixation  of  atmospheric  nitrogen  for  the  production 
of  fertilizers  and  explosives  is  of  vast  importance  to  the  agri- 
cultural and  military  interests. 

Carborundum,  calcium  carbide,  fused  quartz,  silicon, 
graphite,  phosphorus  and  steel  alloys  of  various  sorts  are 
typical  important  electric-furnace  products. 

Immense  quantities  of  caustic  soda  and  chlorine  are  now 


ELECTROLYTIC  CONDUCTION  131 

produced  commercially  from  the  electrolysis  of  sodium 
chloride  solution  in  specially  constructed  cells.  Fig.  52 
shows  one  room  of  a  chlorine  plant  at  Edgewood,  Md., 
which  produced  12J  tons  of  gaseous  chlorine  per  day.  It 
may  safely  be  said  that  the  electrochemical  industries  will 
expand  rapidly  within  the  next  decade;  in  fact,  their  growth 
is  limited  only  to  the  extent  of  future  electric -power  de- 
velopment. 


SUMMARY    OF   CHAPTER  V 

AN  ELECTRIC  CURRENT  FLOWS  THROUGH  A  LIQUID 
because  there  are  ions  present  in  the  liquid.  Such  a  liquid  is 
called  an  ELECTROLYTE. 

A  NEGATIVE  ION  is  an  atom  with  a  greater  number  of 
electrons  than  normal  attached. 

A  POSITIVE  ION  is  one  deficient  in  electrons. 

AN  ELECTRON  is  a  particle  of  negative  electricity,  the 
smallest  particle  into  which  it  is  possible  to  divide  electricity. 

A  CURRENT  OF  ELECTRICITY  really  consists  of  a  flow 
of  these  negative  particles.  The  direction  of  this  flow  is  the 
reverse  of  that  which  we  are  accustomed  to  call  the  positive 
direction  of  current  flow. 

ACIDS  AND  SALTS  IN  SOLUTION  break  up  into  positive 
and  negative  ions.  The  positive  ions  move  through  the  liquid 
to  the  cathode  or  plate  at  which  the  electric  current  leaves  the 
solution.  The  negative  ions  move  in  the  opposite  direction. 

THE  CONDUCTIVITY  OF  A  LIQUID  is  proportional  to  the 
number  of  ions  present  and  their  mobility. 

A  DIFFERENCE  OF  ELECTRICAL  POTENTIAL  is  set  up 
between  two  different  metals  when  they  are  dipped  into  an 
electrolyte  which  is  capable  of  combining  chemically  with 
one  of  them.  The  amount  of  this  potential  difference  depends 
solely  upon  what  metals  are  used. 

THE  ELECTRIC  POTENTIAL  SERIES  OF  METALS 
is  an  arrangement  based  on  the  voltages  which  exist  between 
the  various  metals  if  one  terminal  is  hydrogen. 

FARADAY'S  LAW  states  that  the  weight  of  an  element  lib- 
erated from  a  solution  can  be  found  by  the  equation 

wQ 


where 


m  =  the  weight  in  grams  liberated, 
w  =  the  atomic  weight  of  the  substance, 
Q  =  the  coulombs  of  electricity  passed, 
n  =  the  valence  of  the  element. 
132 


ELECTROLYTIC  CONDUCTION  133 

THE  ELECTROCHEMICAL  EQUIVALENT  of  an  element  is 
the  constant 


and  is  the  weight  in  grams  deposited  by  one  coulomb. 

ONE  FARADAY   =  96,640  coulombs 

=  6.1  X  1023  electrons, 

which  is  also  the  number  of  atoms  in  a  gram-atom  of  a  sub- 
stance. 

A  PRIMARY  CELL  consists  of  two  different  metals  in  con- 
tact with  an  electrolyte  which  combines  chemically  with  one 
of  them.  The  metal  consumed  is  generally  zinc. 

A  STORAGE  CELL  is  similar  to  a  primary  cell  except  that 
the  chemical  actions  are  entirely  reversible. 

THE  LEAD  STORAGE  CELL  consists  of  lead-peroxide 
positive  plates,  metallic  lead  negative  plates  and  dilute  sul- 
phuric acid  electrolyte. 

THE  EDISON  CELL  has  nickel-peroxide  positive  plates, 
iron  negative  plates  and  21%  solution  of  potassium  hydroxide 
electrolyte. 

ELECTROLYTIC  REFINING  OF  METALS  is  carried  out  on 
a  large  scale  to  secure  pure  copper.  The  process  consists  of 
electro-plating  from  a  solution  of  copper  sulphate  and  de- 
pends upon  electrochemical  action. 

ELECTROLYSIS  DOES  DAMAGE  to  water  pipes  and  re- 
inforcing rods  in  concrete  where  any  stray  currents  in  leaving 
the  pipes  or  rods  flow  through  some  electrolyte  surrounding 
the  pipes  or  rods. 

A  LARGE  NUMBER  OF  CHEMICAL  PROCESSES  are 
now  conducted  by  the  aid  of  electricity.  Aluminum,  oxygen, 
caustic  soda  and  chlorine  are  a  few  of  the  more  common  sub- 
stances commercially  produced  by  electrolytic  action. 


PROBLEMS   ON   CHAPTER  V 

Prob.  21-5.  If  the  internal  resistance  of  a  bivalent  nickel 
plating  bath  is  0.5  ohm,  and  1  volt  is  applied,  what  energy  in 
kilowatt-hours  is  used  in  plating  one  pound  of  nickel? 

Prob.  22-5.  If  the  voltage  in  the  above  problem  is  increased 
to  2  volts,  what  will  be  the  new  amount  of  energy  per  pound? 

Prob.  23-5.  Ten  thousand  articles  each  with  a  surface  area 
of  1  square  foot  are  to  be  plated  with  0.003-inch  thick  nickel 
coat.  Take  the  density  of  nickel  as  8.6  grams  per  cubic  centi- 
meter. If  1  volt  is  used  in  the  solution  of  Prob.  21-5  and  the 
energy  cost  is  20^  per  kilowatt-hour,  what  is  the  energy  cost 
per  article? 

Prob.  24-5.  If  the  labor  cost  for  plating  bath  attendance  is 
4^  per  article  under  the  conditions  of  Prob.  23-5,  and  this  can 
be  reduced  to  3^  if  the  time  of  plating  is  halved,  will  it  pay  to 
use  a  2-volt  supply? 

Prob.  25-5.  Hydrogen  under  standard  conditions  of  tem- 
perature and  pressure  weighs  0.09  gram  per  cubic  foot.  How 
long  will  it  take  1000  amperes  to  liberate  enough  hydrogen  by 
electrolytic  decomposition  to  fill  a  balloon  which  is  spherical 
and  40  feet  in  diameter? 

Prob.  26-6.  At  a  voltage  of  1.5  and  at  10j£  per  kilowatt- 
hour,  what  is  the  cost  of  energy  for  filling  the  balloon  of  Prob. 
25-5? 

Prob.  27-5.  Compare  the  energy  costs  of  plating  nickel 
and  silver  (assuming  the  resistance  of  the  plating  bath  to  be  the 
same  in  each  case)  for  equal  weights  deposited.  For  equal 
thicknesses  on  an  article.  (Density  of  silver  is  10.5  grams  per 
cubic  centimeter.) 

Prob.  28-5.  A  dry  cell  delivering  1.5  volts  will  supply  I 
ampere  intermittently  for  a  total  of  50  hours.  The  cell  costs 
30£  new.  What  is  the  cost  of  power  obtained  in  this  way? 

Prob.  29-5.  An  average  per  plug  of  0.04  joule  output  of 
electrical  energy  is  generally  used  by  the  ignition  system  to 
produce  a  proper  spark  at  the  spark  plug  of  a  gasoline  engine. 
With  an  8-cylinder  4-cycle  engine  running  3200  r.p.m.,  what  is 

134 


ELECTROLYTIC  CONDUCTION  135 

the  total  input  in  watts  to  produce  the  sparks,  assuming  an 
efficiency  of  the  ignition  system  of  20%? 

Prob.  30-5.  If  the  engine  of  Prob.  29-5  is  supplied  by  cells 
of  the  type  of  Prob.  28-5,  how  long  will  6  such  cells  last  on 
continuous  running? 

Prob.  31-5.  An  iron  water  pipe  receives  a  stray  current  of 
5  amperes  from  a  trolley  system.  Where  this  current  leaves 
the  pipe,  electrolysis  occurs.  Assuming  Faraday's  law,  how 
long  will  it  take  to  remove  10  pounds  of  the  iron  of  the  pipe? 

Prob.  32-5.  A  copper  plate  is  riveted  to  the  iron  hull  of  a 
ship.  The  salt  sea  water  forms  an  active  electrolyte.  Before 
polarization  what  voltage  acts  and  in  what  direction?  After 
the  cathode  is  covered  with  hydrogen,  what  voltage  acts? 

Prob.  33-5.  The  anode  is  attacked  when  the  current  flows 
in  a  case  such  as  in  Prob.  32-5.  Will  the  presence  of  the  copper 
plate  cause  damage  to  the  iron  plates  of  the  hull?  What  if  the 
copper  plate  is  replaced  by  one  of  zinc? 

Prob.  34-5.  Iron  is  protected  against  corrosion  in  the 
presence  of  moisture  which  may  contain  salts  which  render  it 
active  as  an  electrolyte  by  covering  with  a  coating  of  zinc. 
This  is  done  in  several  ways,  as  in  the  galvanizing  process. 
Tin  is  also  used.  Compare  the  two  as  regards  corrosion  of  the 
iron  if  there  are  pinholes  in  the  coating. 


CHAPTER  VI 
THE  MAGNETIC  CIRCUIT 

Practically  all  electric  power  machinery  depends  for  its 
operation  upon  the  inter-relation  between  electricity  and 
magnetism.  The  generator,  the  motor,  the  transformer, 
contain  not  only  electric  circuits  but  also  magnetic  circuits, 
and  depend  for  their  operation  upon  the  mutual  effects 
produced  between  the  two. 

55.  Relation  Between  Electricity  and  Magnetism.  When 
an  electric  current  flows  along  a  wire,  there  are  evidences 
of  its  presence  not  only  within  the  wire  itself  but  also  in 
the  space  around  it.  Inside  the  conductor,  the  electric  cur- 
rent produces  heating  and  chemical  effects,  but  it  produces 
magnetic  effects  in  the  space  outside  of  the  wire  as  well 
as  within  it.  As  early  as  1820,  Oersted  discovered  that  a 
compass  needle  in  the  vicinity  of  a  wire  carrying  current 
was  deflected,  and  Ampere,  upon  hearing  of  this  result, 
soon  worked  out  the  law  underlying  this  effect.  This  law 
bears  his  name.  It  is  principally  to  Faraday,  however, 
that  we  owe  the  complete  examination  of  the  quantitative 
relationship  between  electric  current  and  magnetism,  for 
it  was  due  to  his  painstaking  experimentation  and  clear 
logic  that  most  of  the  laws  underlying  present-day  electrical 
engineering  were  discovered  and  put  into  form. 

Without  the  knowledge  of  these  various  relations  be- 
tween electric  and  magnetic  circuits,  there  would  be  no 
electrical  engineering  to-day.  The  physicist  would  have  at 
his  command  only  voltaic  cells  and  weak  permanent  magnets. 
They  would  be  largely  toys,  and  powerful  electrical  machinery 
could  not  exist.  It  is  only  by  the  use  of  coils  carrying  electric 
current  that  powerful  magnetic  fields  can  be  established, 

136 


THE  MAGNETIC  CIRCUIT 


137 


such  for  instance  as  in  the  large  lifting  magnets  which  raise 
twenty  tons  at  a  load.  By  the  use  of  these  powerful  fields, 
we  can  in  turn  generate  large  voltages  and  currents,  that  is, 

Metal  Wire 


Current 


Source  of 

Electromotive 

Force 


Battery 


FIG.  53. 


The  electromotive  force  of  the  battery  causes  a  current  to 
flow  around  the  metallic  circuit. 


large  amounts  of  electric  power,  making  possible  the  50,000- 
kilowatt  turbo  generator  as  we  know  it  to-day.  On  the 
other  hand,  even  such  a  delicate  instrument  as  the  telephone 
receiver  depends  for  its  action  upon  electromagnetic  effects. 


Iron  Core 


Coil- 


Magnetic  Flux 


•  Source  of 
Magnetomotive 
Force 


FIG.  54.     The  magnetomotive  force  of  the  coil  sets  up  a  magnetic 
flux  in  the  iron  circuit. 

56.  The  Magnetic  Circuit.  Just  as  electricity  or  electric 
current  is  caused  to  flow  in  an  electric  circuit,  so  magnetism 
or  magnetic  flux  can  be  set  up  in  a  magnetic  circuit.  In 
Fig.  53,  the  battery  acts  as  a  source  of  electromotive  force 


138      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

and  forces  an  electric  current  through  an  electric  circuit 
consisting  of  a  metal  wire.  Similarly  in  Fig.  54,  a  coil 
carrying  an  electric  current  can  be  caused  to  act  as  a  source 
of  magnetomotive  force,  and  to  force  magnetic  flux  through 
an  iron  core  which  constitutes  a  magnetic  circuit. 

It  will  be  seen  that  in  many  ways  these  two  circuits  are 
similar  and  it  is  found  that  the  laws  governing  them  may  be 
written  in  much  the  same  form.  There  are,  however,  im- 
portant differences,  principal  among  which  differences  is 
the  following.  In  the  electric  circuit,  the  current  produces 
an  effect  in  the  wire  such  as  heating,  even  when  the  current 
is  absolutely  steady.  A  magnetic  flux,  however,  produces 
its  principal  effect  when  it  is  varying  in  the  magnetic  circuit, 
and  in  general  does  not  make  its  presence  felt  when  it  is 
steady.  A  magnetic  circuit  may  carry  a  steady  magnetic 
flux  indefinitely  without  producing  any  heating  of  the 
circuit,  and  consequently  without  any  expenditure  of  energy, 
at  least  so  far  as  the  magnetic  circuit  itself  is  concerned. 
There  are  other  important  differences  which  will  be  brought 
out  below. 


Ammeter^. 

&5* 

r 

Iron, 

\ 

Voltmeters 

^m$/ 

o^  

O    O- 

c~       "~~i 

-O      O 

_L         Switch 

( 

Magnetizing            < 

^' 

2-f 

L.  .   >        AAAAAJL 

i 

''Coil  (A)                  ( 

-^~ 

1 

35 

> 

X 

J 

rv.ii  (K\ 

FIG.  55.     A  change  of  current  strength  in  coil  A  will  set  up  a  voltage 
in  coil  B  and  cause  the  voltmeter  to  deflect. 

57.  Measurement  of  Flux.  A  magnetic  circuit  may  be 
constructed  by  making  a  closed  iron  core  and  winding  upon 
it  a  coil  of  wire.  When  a  current  is  passed  through  this 
coil,  as  for  instance  by  using  a  battery  as  shown  in  Fig. 
55,  magnetic  flux  will  be  set  up  in  the  iron.  The  presence 
of  this  flux  may  be  made  apparent  in  several  ways,  but  it 
will  be  convenient  for  us  to  examine  at  this  time  one  way 


THE  MAGNETIC  CIRCUIT  139 

which  is  of  principal  importance.  If  we  wind  a  second  coil 
of  wire  on  this  same  core  and  connect  it  to  a  voltmeter, 
with  the  zero  position  at  the  middle  of  the  scale,  the  fol- 
lowing effects  may  be  observed.  Although  the  two  coils 
have  no  electrical  connection  whatever  with  each  other, 
yet  when  the  switch  in  coil  A  is  closed,  the  voltmeter  con- 
nected to  coil  B  will  suddenly  deflect,  thus  showing  that  a 
voltage  in  a  certain  direction  has  been  set  up  in  the  meas- 
uring coil.  Since  these  two  coils  have  no  electrical  connec- 
tion, it  is  reasonable  to  suppose  that  the  flux  set  up  by  the 
current  in  coil  A  has  something  to  do  with  producing  the 
voltage  set  up  in  coil  B.  When,  however,  a  current  is 
flowing  steadily  in  the  magnetizing  coil  A,  the  voltmeter 
in  coil  B  will  return  to  its  zero  position,  and  there  will  now 
be  no  indication  of  any  effect  in  the  core.  If,  however, 
the  switch  in  the  circuit  of  coil  A  is  opened,  the  voltmeter 
will  suddenly  deflect  in  the  opposite  direction,  thus  showing 
voltage  in  the  opposite  direction  from  that  of  the  first 
voltage.  It  thus  appears  that  the  flux  produces  an  effect, 
namely,  introduces  a  voltage  into  the  measuring  coil,  while 
it  is  in  the  process  of  changing,  —  a  voltage  in  one  direction 
when  flux  is  being  set  up  and  a  voltage  in  the  opposite 
direction  when  it  is  decreasing  to  zero. 

We  may  observe  these  effects  more  exactly  by  keeping 
the  battery  switch  closed  and  utilizing  "a  rheostat  and  an 
ammeter  in  the  battery  circuit.  We  shall  find  that  when 
the  ammeter  reads  steadily  at  any  position,  the  voltmeter 
will  remain  at  the  center  of  its  scale.  When  the  ammeter 
shows  that  the  current  is  increasing,  the  voltmeter  will 
deflect  in  one  direction;  and  when  the  ammeter  shows  that 
the  current  is  decreasing,  the  voltmeter  will  deflect  in  the 
opposite  direction.  By  operating  the  rheostat  at  various 
rates,  we  can  make  the  current  change  quickly  or  slowly. 
It  will  be  found  that  the  voltage  produced  will  be  exactly 
proportional  to  the  rate  at  which  the  flux  in  the  magnetic 
circuit  changes.  This  is  the  same,  except  for  certain  effects 


140      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

to  be  noted  below,  as  the  rate  at  which  the  current  in  the 
magnetizing  coil  changes.  It  is  natural,  therefore,  for  us 
to  use  the  voltage  thus  produced  in  a  measuring  coil  to 
measure  the  amount  of  magnetic  flux  produced  in  the  iron 
magnetic  circuit  by  the  magnetizing  coil  around  it. 

Since  it  is  hard  to  make  accurate  measurements  of  the 
rate  at  which  the  current  is  varying  at  given  instants  as 
outlined  above,  it  is  more  convenient  to  use  a  ballistic 
galvanometer  for  measuring  the  amount  of  flux  in  the 
magnetic  circuit.  The  ballistic  galvanometer  is  simply  a 
galvanometer  with  a  very  long  period.  If  a  voltage  is 
applied  to  such  an  instrument  for  a  short  interval  and  then 
removed,  the  deflection  of  the  galvanometer  will  be  pro- 
portional to  the  product  of  the  voltage  applied  and  the 
length  of  time  that  it  is  connected.  The  reason  for  this 
effect  will  be  more  apparent  when  we  have  studied  the 
theory  underlying  electrical  instruments,  but  for  the  present 
we  may  simply  assume  that  the  ballistic  galvanometer  can 
be  used  to  measure  in  volt-seconds  the  amount  of  a  sudden 
impulse  applied  to  it. 

If  the  voltage  applied  to  a  ballistic  galvanometer  is 
varying  during  the  short  interval  that  it  is  applied,  the  de- 
flection of  the  galvanometer  will  be  proportional  to  the  in- 
tegral of  the  voltage  taken  over  the  interval  of  time;  that  is, 


kife  dt, 


(1) 


where  d  =  the  deflection  of  the  galvanometer,  t  =  the  time 
in  seconds  and  ki  =  the  constant  of  the  galvanometer. 
We  have  seen  above  that  the  voltage  produced  by  the 
changing  flux  in  the  magnetic  circuit  is  proportional  to  the 
rate  at  which  the  flux  changes;  that  is, 


where  <j>  =  the   magnetic   flux  in  maxwells  and  k2  =  the 
proportionality  factor.     Then  the  deflection  of  a  ballistic 


THE  MAGNETIC  CIRCUIT  141 

galvanometer  connected  to  a  measuring  coil  about  a  mag- 
netic circuit  will  be  proportional  to 

*-/*$*  (3> 

or  simply  • 

d  =  k<t>,  (4) 

where  k  is  a  constant  combining  k\  and  k%.  That  is,  the 
galvanometer  deflection  is  a  measure  of  the  total  amount 
by  which  the  flux  changes  in  the  magnetic  circuit.  Thus, 
if  we  connect  a  ballistic  galvanometer  as  shown  in  Fig.  56, 
putting  the  galvanometer  in  place  of  the  voltmeter,  and 
suddenly  change  the  amount  of  flux  in  the  magnetic  circuit 
by  varying  the  battery,  the  deflection  of  the  galvanometer 
will  be  proportional  to  the  total  amount  by  which  the  flux 
is  changed.  The  deflection  which  occurs  when  the  flux  in 
the  circuit  is  first  set  up  may  be  taken  as  the  measurement 
of  the  amount  of  flux  in  the  circuit. 

As  the  c.g.s.  unit  of  magnetic  flux  we  take  that  amount 
of  flux  which,  when  established  in  the  magnetic  circuit, 
will  produce  one  abvolt-second  in  a  coil  of  one  turn  wound 
on  the  magnetic  circuit,  and  we  call  this  unit  of  flux  a  max- 
well. The  number  of  maxwells  of  flux  in  a  magnetic  cir- 
cuit may  thus  be  measured  by  measuring  the  abvolt-sec- 
onds  indicated  by  a  ballistic  galvanometer  connected  to  a 
measuring  coil  around  a  circuit  when  the  flux  is  set  up  or 
decreased  to  zero,  the  measuring  coil  being  of  one  turn.  It 
is  more  convenient  in  practice  to  use  a  measuring  coil  of 
several  turns  (an  equal  voltage  being  set  up  in  each  turn,  of 
course)  and  divide  the  resulting  deflection  of  the  galva- 
nometer by  the  number  of  turns  used.  It  is  also  better,  for 
reasons  explained  below,  to  use  an  arrangement  such  as  is 
shown  in  Fig.  56  and  reverse  the  magnetizing  current  by 
means  of  a  reversing  switch.  The  deflection  of  the  gal- 
vanometer when  the  reversing  switch  is  thrown  from  one 
side  to  the  other  is  thus  due  to  a  change  of  flux  from  a  maxi- 


142      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


mum  in  one  direction  to  a  maximum  in  the  other  direction. 
The  amount  of  flux  as  computed  from  the  galvanometer 
deflection  should  be  divided  by  two,  in  order  to  give  a  cor- 


—  — 
—     - 

^ 

Ballistic 
Galvanometer 

^Coil                     ] 

* 

Measuring 
Coil 

V 

FIG.  56.     Reversing  the  switch  in  the  circuit  of  the  magnetizing  coil 
causes  a  voltage  to  be  set  up  in  the  measuring  coil. 

rect  result  for  the  maximum  amount  of  flux  in  the  magnetic 
circuit. 

Prob.  1-6.  The  flux  through  a  coil  of  one  turn  is  increased 
gradually  from  zero  to  150,000  maxwells  in  two  seconds.  What 
is  the  average  voltage  produced  in  abvolts?  In  volts? 

Prob.  2-6.  If  the  coil  of  Prob.  1-6  has  200  turns,  what 
average  voltage  will  be  produced? 

Prob.  3-6.  If  the  voltmeter  in  Fig.  55  shows  a  voltage  of 
2  millivolts,  and  the  coil  to  which  it  is  connected  has  1000 
turns,  at  what  rate  is  the  flux  changing  in  the  iron  core? 

58.  Flux  Lines.  Euler,  in  his  "  Letters  to  a  German 
Princess,"  written  in  1761,  describes  a  magnet  and  speaks 
of  the  "  lines  of  flow  of  the  magnetic  fluid."  It  was  from 
this  sort  of  description  of  the  magnetic  effect  in  a  magnetic 
circuit  that  we  have  obtained  the  terms  "  magnetic  flux  " 
and  "  magnetic  flux  lines." 

Just  as  it  is  convenient  for  us  to  map  out  the  manner  in 
which  water  flows  through  a  nozzle  (Fig.  57)  by  drawing 
lines  to  indicate  the  direction  which  the  water  takes  at 
every  point,  so  it  is  convenient  for  us  to  represent  the  path 
which  the  magnetic  flux  takes  in  a  magnetic  circuit  by  means 
of  lines  showing  at  every  point  its  direction,  (Fig.  58). 
We  may  also  show  the  direction  which  an  electric  current 
takes  at  every  point  in  an  electric  circuit  by  means  of  lines 
similarly  drawn. 


THE  MAGNETIC  CIRCUIT 


143 


Faraday  was  the  first  to  utilize  this  manner  of  describing 
the  path  taken  by  magnetic  flux  in  computations,  and  to 
show  the  great  benefit  of  thus  representing  magnetic  flux. 


FIG.  57.  The  lines  show  the 
amount  and  direction  of 
water  flowing  from  an  orifice. 


FIG.  58.  The  lines  show  the 
amount  and  direction  of  mag- 
netic flux. 


Suppose  that  in  Fig.  57,  in  drawing  the  lines  representing 
the  flow  of  water  through  an  orifice,  we  observe  the  following 
arrangement.  Draw  one  line  for  each  cubic  inch  per  second 
of  water  flowing.  That  is,  consider  the  stream  of  water  to 
be  made  up  of  a  large  number  of  small  streams  flowing 
together  and  each  carrying  one  cubic  inch  of  water  per 
second.  Represent  each  of  these  unit  streams  by  one  line. 
The  lines  will  then  show  not  only  the  direction  that  the 
water  takes,  but  also  the  amount  of  water  flowing.  The 
closer  together  these  lines  are  situated,  the  more  rapidly 
will  the  water  be  flowing  at  that  point.  The  density  of 
the  lines,  that  is,  the  number  of  lines  per  square  inch  passing 
through  a  cross-section  perpendicular  to  the  stream,  will, 
in  fact,  show  the  velocity  of  the  water  at  any  point  in  inches 
per  second. 

Similarly,  in  representing  the  magnetic  flux  in  a  magnetic 
circuit,  we  draw  one  line  to  represent  a  flux  of  one  maxwell; 
in  fact,  we  often  abbreviate  by  saying  so  many  lines  of  flux 
instead  of  saying  so  many  maxwells  of  flux.  The  number 
of  lines  per  square  inch  of  cross-section,  that  is,  the  density 
of  the  lines,  we  take  as  a  measurement  of  the  intensity  of 
the  magnetic  effect  produced.  A  flux  density  of  one  max- 


144      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

well  (or  one  line)  per  square  centimeter  of  cross-section  is 
called  a  flux  density  of  one  gauss.  In  ordinary  magnetic 
circuits  constructed  of  iron,  we  employ  flux  densities  of 
many  thousands  of  gausses. 

The  lines  thus  used  to  describe  the  direction  and  intensity 
of  magnetic  flux  do  not  have  any  physical  existence,  any 
more  than  do  the  stream  lines  describing  the  flow  of  water. 
They  are  simply  a  convenient  means  of  describing  the 
magnetic  effect. 

When  electric  current  flows  in  an  electric  circuit,  there  is 
an  actual  transfer  of  electrons  along  the  circuit.  When 
magnetic  flux  flows  in  a  magnetic  circuit,  on  the  other  hand, 
we  do  not  now  believe  that  there  is  any  actual  flow  of  any- 
thing. This  is  indicated  by  the  fact  noted  above,  that 
although  an  electric  current  produces  heating  of  the  wire 


FIG.  59.     The  lines  show  the  strains  at  various  parts  of  the  beam. 

while  it  flows,  magnetic  flux,  as  long  as  it  is  steady,  produces 
no  heat  or  chemical  effect  in  the  iron  through  which  it 
passes.  The  magnetic  flux  is  more  nearly  in  the  nature 
of  a  strain  set  up  in  the  material.  The  flux  lines  should  be 
considered  as  being  lines  of  strain  due  to  stress.  We  might 
similarly  indicate  the  direction  and  intensity  of  the  strains 
in  a  beam  under  load  as  shown  in  Fig.  59,  the  horizontal 
lines  representing  the  direction  of  the  strains  at  the  various 
points. 

It  is  important  to  note  that  magnetic-flux  lines  are  always 


THE  MAGNETIC  CIRCUIT  145 

continuous.  In  other  words,  a  magnetic-flux  line  can  never 
have  an  end.  This  is  true  of  the  flow  lines  in  a  stream  of 
water.  For  a  flow  line  to  end  would  mean  that  a  certain 
quantity  of  water  per  second  disappeared  entirely  at  some 
point  in  the  stream.  In  the  same  way,  the  magnetic- 
flux  lines  are  always  closed  lines,  and  go  completely  around 
the  magnetic  circuit  whatever  its  shape  may  be. 

59.  Magnetomotive  Force.  A  voltaic  battery  intro- 
uced  into  an  electric  circuit  provides  an  electromotive  force 
which  forces  current  around  the  circuit.  Similarly  a  coil 
of  wire  carrying  current  produces  a  magnetomotive  force 
which  may  force  a  magnetic  flux  around  a  magnetic  circuit. 
We  should  note,  however,  one  important  difference.  The 
voltaic  battery  is  an  integral  part  of  the  electric  circuit, 
and  produces  its  electromotive  force  right  in  the  circuit 
itself  by  means  of  chemical  action.  On  the  other  hand,  a 
magnetomotive  force  in  a  magnetic  circuit  may  be  supplied 
by  a  coil  of  wire  carrying  current  even  though  the  coil  does 
not  physically  touch  the  magnetic  circuit  at  all.  It  is 
merely  necessary  that  the  coil  of  wire  surround  one  part 
of  the  magnetic  circuit,  that  is,  that  the  coil  and  the  mag- 
netic circuit  interlink. 

The  total  magnetic  flux,  that  is,  the  number  of  magnetic 
lines  produced  in  a  magnetic  circuit,  has  been  found  ex- 
perimentally to  be  proportional  to  the  current  in  the  mag- 
netizing coil  when  the  magnetic  circuit  does  not  contain 
iron  or  other  magnetic  material.  The  flux  has  also  been 
found  to  be  proportional  to  the  number  of  turns  of  wire 
used  in  the  magnetizing  coil,  that  is,  to  the  number  of  times 
the  current  in  the  coil  links  the  magnetic  circuit.  The 
magnetomotive  force  of  a  magnetizing  coil  is  hence  propor- 
tional to  the  product  of  the  number  of  turns  in  the  coil  and 
the  electric  current  flowing.  This  assumes  that  the  turns  are 
all  wound  in  the  same  direction.  If  a  number  of  turns  are 
wound  on  in  the  reverse  direction,  their  effect  will  cancel 
the  effect  of  an  equal  number  of  turns  correctly  wound. 


146      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

The  magnetomotive  force  is  hence  proportional  to  the 
product  of  the  electric  current  and  the  number  of  linkages 
between  the  wire  and  the  magnetic  circuit. 

60.  Reluctance.     The  current  flowing  in  an  electric  cir- 
cuit is  proportional  to  the  electromotive  force  and  inversely 
proportional  to  the  resistance.     This  is  Ohm's  law.     Sim- 
ilarly, the  amount  of  flux  in  a  magnetic  circuit  is  propor- 
tional to  the  magnetomotive  force  applied  and  inversely  pro- 
portional to  what  is  called  the  reluctance  of  the  circuit. 

The  resistance  of  a  uniform  electric  circuit  is  found  ex- 
perimentally to  be  proportional  to  the  length  of  the  circuit 
and  inversely  proportional  to  the  cross-section.  The  pro- 
portionality factor  is  called  the  resistivity  of  the  material 
used,  and  its  reciprocal  the  conductivity  of  the  material. 
Similarly,  the  reluctance  of  a  magnetic  circuit  is  proportional 
to  the  length  of  the  circuit  and  inversely  proportional  to  its 
cross-section.  Analogous  to  the  conductivity  of  an  electric 
circuit  we  have  the  permeability  of  a  magnetic  circuit.  The 
conductivity  depends  upon  the  material  used  for  making 
an  electric  circuit,  its  temperature  and  sometimes  other 
effects.  Similarly,  the  permeability  of  a  magnetic  circuit 
depends  upon  the  material  used  and  upon  other  very  impor- 
tant factors  to  be  outlined  below.  The  conductivity  of 
a  given  material  is  denned  as  the  conductance  (that  is, 
the  reciprocal  of  the  resistance)  of  a  unit  cube  of  the 
material  used  in  the  electric  circuit.  In  the  same  way, 
the  permeability  of  a  given  material  is  the  reciprocal  of 
the  reluctance  of  a  unit  cube  of  the  material  of  a  magnetic 
circuit. 

The  unit  of  resistance  is  the  ohm.  The  unit  of  reluctance 
is  the  oersted. 

61.  Ohm's  Law  for  Magnetic  Circuits.    For  the  electric 
circuit,  we  write 

electromotive  force 

current  = r- • 

resistance 


THE  MAGNETIC  CIRCUIT  147 

This  may  be  abbreviated  to 

T       E 
I  =  R' 

where,  if  we  are  using  the  c.g.s.  system  of  units,  I  is  the 
current  in  abamperes,  E  is  the  electromotive  force  in  abvolts 
and  R  is  the  resistance  in  abohms.  In  the  practical  system 
of  units,  I  is  the  current  in  amperes,  E  is  the  electromotive 
force  in  volts  and  R  is  the  resistance  in  ohms.  The  resist- 
ance, we  have  seen,  may  be  written 


where  I  is  the  length  and  A  is  the  cross-section  of  the  mag- 
netic circuit  and  7  is  the  conductivity  of  the  material  used. 
Similarly,  for  the  magnetic  circuit,  we  may  write 

magnetomotive  force 
flux  =  —  -  -  9 

reluctance 

which  may  be  abbreviated  to 

*-|.  .  (5) 

where  <£  is  the  flux  in  maxwells,  £F  is  the  magnetomotive 
force  in  gilberts  and  91  is  the  reluctance  in  oersteds.  These 
units  are  all  in  the  c.g.s.  system.  In  the  same  way,  the 
reluctance  may  be  written 


where  I  is  the  length  and  A  the  cross-section  of  the  magnetic 
circuit  and  /z  is  the  permeability  of  the  material  used. 

We  saw  above  that  the  magnetomotive  force  was  propor- 
tional to  the  current  and  to  the  number  of  turns  in  the  mag- 
netizing coil. 

Magnetomotive  force,  however,  is  not  simply  NI,  where 


148      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

N  is  the  number  of  turns,  but  we  define  the  magnetomotive 
force  as 


/ 

where  fFis  the  magnetomotive  force  in  gilberts,  N  is  the 
number  of  turns  and  /  is  the  current  in  abamperes.  If 
the  current  is  expressed  in  amperes,  the  magnetomotive 
force  in  gilberts  will  be 

7=  fi*NL  (7) 

The  proportionality  factor,  4ir,  is  introduced  in  this 
manner  in  order  to  make  the  values  of  permeability  come 
out  conveniently.  We  shall  see  below  that  if  we  define 
the  magnetomotive  force  in  the  above  manner,  we  can  show 
that  the  permeability  of  air  comes  out  unity.  This  is,  of 
course,  convenient  in  calculation.  In  selecting  electrical 
units,  we  might  have  chosen  a  different-sized  unit  for  the 
volt  if  we  had  wished;  and  if  we  had  chosen  correctly,  the 
resistivity  of  copper  might  have  been  made  to  come  out  as 
unity.  It  would  not  have  been  worth  while  to  do  this  with 
electrical  units,  for  many  different  materials  are  used  as 
conductors  and  their  resistivities  vary  greatly.  On  the 
other  hand,  the  permeabilities  of  almost  all  materials  are 
practically  identical  with  that  of  air,  namely  unity.  There 
are  only  a  few  magnetic  materials  which  have  permeabilities 
differing  greatly  from  unity;  iron,  the  principal" material 
used  in  practice,  may  have  permeabilities  of  several  thousand. 
It  is  convenient  to  have  our  unit  of  magnetomotive  force, 
the  gilbert,  so  chosen  that  the  permeability  of  nearly  all 
materials  is  almost  exactly  unity. 

The  flux  flowing  in  a  magnetic  circuit  which  is  interlinked 
by  a  magnetizing  coil  of  a  certain  number  of  turns,  N,  and 
carrying  a  certain  current,  I  abamperes,  can  accordingly 
be  found  from  the  formula 

4irNI . 


THE  MAGNETIC  CIRCUIT  149 

or  if  the  current  is  expressed  in  amperes  instead  of  abamperes, 

by 

(8) 


The  flux  density,  that  is,  the  intensity  of  magnetization 
produced,  can  be  found  by  dividing  the  total  flux  by  the 
cross-section.  This  flux  density  is  thus  expressed  in  lines 
per  square  centimeter  or  lines  per  square  inch.  The  den- 
sity in  lines  per  square  centimeter  is  also  called  the  density 
in  gausses.  The  symbol  B  is  always  used  to  denote  flux 
density.  Thus 

0       4<irNI 
B  =      =  —  -  gausses. 


/* 
This  last  formula  may  be  written  in  the  form 

B  =  — j —  n  gausses.  (9) 

The  quantity  —, —  or  j 

is  the  magnetomotive  force  per  unit  of  length  of  the  circuit. 
It  is  thus  expressed  in  gilberts  per  centimeter.  It  is  analo- 
gous evidently  to  the  potential  gradient  along  an  electric  cir- 
cuit. This  quantity,  called  the  magnetizing  force,  has  re- 
ceived the  letter  H  to  designate  it.  We  thus  have 

B  =  »H;  (10) 

that  is,  the  flux  density  is  equal  to  the  magnetizing  force 
times  the  permeability.  This  is  analogous  to  the  expression 
for  the  electric  circuit  that  the  current  density  is  equal  to 
the  conductivity  times  the  potential  gradient.  We  shall 
make  much  use  of  this  expression  later. 


150      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  4-6.  The  magnetizing  coil  in  Fig.  55  has  40  turns 
and  the  ammeter  shows  a  current  of  2  amperes.  What  magneto- 
motive force  is  acting  on  the  magnetic  circuit? 

Prob.  5-6.  If  the  core  in  Prob.  4-6  is  of  annealed  steel  of 
permeability  2500,  4  square  inches  in  cross-section,  and  of  20 
inches  total  length,  what  is  the  reluctance  of  the  magnetic  circuit? 

Prob.  6-6.  What  flux  will  exist  in  the  core  under  the  con- 
ditions of  Prob.  5-6? 

Prob.  7-6.  What  is  the  magnetizing  force  and  what  is  the 
flux  density  in  Prob.  6-6? 

Prob.  8-6.  A  cast-iron  ring  of  square  cross-section,  3  inches 
inside  diameter,  4  inches  outside  diameter,  and  0.5  inch  thick, 
is  wound  with  a  coil  of  200  turns.  When  a  current  of  2.7  amperes 
is  passed  through  this  coil,  the  ring  carries  a  total  flux  of  8000 
maxwells.  (a)  What  is  the  permeability  of  the  cast  iron? 
(6)  What  is  the  average  magnetizing  force? 

62.  Reluctances  in  Series  and  in  Parallel.  When  a 
magnetic  circuit  is  uniform,  its  reluctance  may  be  im- 
mediately computed,  provided  its  permeability  is  known, 
from  the  expression 

91  =  —  r  oersteds.  (6) 

fj.A 

If  the  circuit  is  not  uniform,  however,  we  may  still  compute 
its  reluctance,  provided  we  know  the  permeability  of  each 
part,  by  the  same  process  as  was  used  in  computing  the 
resistance  of  a  non-uniform  electric  circuit. 

If  two  reluctances  are  connected  in  series,  the  total  re- 
luctance is  the  sum  of  the  individual  reluctances.  Thus, 
in  Fig.  60,  the  total  reluctance  of  the  magnetic  circuit  shown 
is 


The  same  method  may  be  used  for  any  other  combination 
of  series  reluctances  we  may  wish  to  consider. 
For  two  reluctances  in  parallel,  the  total  reluctance  is, 


THE  MAGNETIC  CIRCUIT 


151 


similarly,  the  reciprocal  of  the  sum  of  the  reciprocals  of  the 
separate  reluctances;  that  is, 

1 


91  = 


(12) 


For  the  magnetic  circuit  shown  in  Fig.  79a,  constructed  of 
two  concentric  iron  rings  of  dimensions  and  permeabilities 
indicated,  the  reluctance  of  the  combination  is  found  thus: 

For  the  inner  ring 

27T4 


1000  X  3  X  1.4 


For  the  outer  ring 


=  0.00599  oersted  ; 


=  0.00874  oersted; 


2000  X  3  X  0.6 
For  the  two  rings  in  parallel 

&  =  xJ^rr  =  0.0035  oersted. 


Permeability  ju  {  -\ -^Permeability  /*3 

FIG.  60.     A  magnetic  circuit  made  of  different  materials  and  having 
a  non-uniform  cross-section. 

Prob.  9-6.     The  soft-steel  piece  of  Fig.  60  has  the  following 
dimensions:     r:  =  T^  inch,  r2  =  |  inch,  a  =  J  inch,   R  =  lj 


152      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

inches.  The  permeability  of  the  part  of  smaller  cross-section 
is  500  and  of  the  larger  cross-section  part  1100.  What  is  the 
total  reluctance  of  this  magnetic  circuit? 

Prob.  10-6.  A  magnetic  circuit  is  composed  of  two  parts  in 
series.  One  part  is  of  cast  steel  with  a  permeability  of  1000 
and  has  the  following  dimensions:  total  length,  21  inches; 
cross-section,  6  square  inches.  The  other  part  is  an  air  gap 
TV  inch  long  and  of  8  square  inches  cross-section.  Find  the  re- 
luctance of  (a)  the  steel  part  of  the  circuit;  (6)  the  air  gap; 
(c)  the  complete  circuit. 

Prob.  11-6.  If  the  air  gap  of  Prob.  10-6  were  put  in  parallel 
with  the  steel  circuit,  what  would  be  the  reluctance  of  the  com- 
bination? Assume  the  permeability  to  remain  constant. 

Prob.  12-6.  Make  a  sketch  of  each  of  the  magnetic  circuits 
in  Prob.  10-6  and  11-6,  marking  the  dimensions  of  parts  and 
showing  flux  paths  in  each. 

Prob.  13-6.  If  the  flux  density  in  the  steel  path  in  Prob. 
11-6  is  9000  gausses,  how  many  maxwells  are  there  in  the  air 
gap? 

63.  Variation  of  Permeability.  In  solving  an  electric 
circuit,  at  least  one  made  in  the  ordinary  manner  of  metal 
wire,  it  is  found  that  the  resistance  to  be  used  is  strictly 
constant,  as  long  as  the  temperature  of  the  circuit  remains 
unchanged.  If  the  temperature  varies,  we  discover  that 
the  resistance  also  varies  with  it,  but  at  any  fixed  tempera- 
ture the  resistance  is  constant.  If,  however,  we  attempt 
to  solve  an  electric  circuit  including  an  electric  arc,  we  soon 
find  that  Ohm's  law  in  its  simple  form  does  not  altogether 
hold.  The  resistance  of  the  arc  itself  is  not  a  fixed  value 
but  depends  upon  the  amount  of  current  flowing  through  it. 

In  the  study  of  magnetic  circuits,  we  are,  in  a  similar  way, 
soon  impressed  with  the  fact  that  the  permeability  of  iron 
is  not  a  fixed  quantity,  but  depends  upon  the  flux  density 
at  which  the  permeability  is  measured.  This  relation  be- 
tween flux  density  and  permeability  is  not  a  formal  one, 
but  is  experimental  only,  and  hence  is  determined  by  actual 
tests  on  the  iron. 

In  Fig.  62  is  shown  a  set  of  curves  which  give  the  relation 


THE  MAGNETIC  CIRCUIT 


153 


154      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


•mo  jad  g>J3qiip  JJ 
sassnBQ  a 


THE  MAGNETIC  CIRCUIT  155 

between  the  permeability  and  the  flux  density  of  various 
materials.  It  will  be  noted  that  at  low  flux  density  the 
permeability  of  iron  is  fairly  low,  that  there  is  a  certain 
flux  density  at  which  maximum  permeability  is  reached, 
and  that  above  this  flux  density  the  permeability  is  decreas- 
ing. It  is  somewhat  more  convenient  for  work  of  com- 
putation to  plot  immediately  the  flux  density  against  the 
magnetizing  force.  Such  a  curve  is  called  the  B-H  curve 
of  the  iron  used.  In  Fig.  61  is  shown  a  set  of  curves  thus 
plotted  for  various  kinds  of  iron  such  as  are  usually  used 
in  magnetic  circuits,  the  curves  giving  directly  the  flux 
density  in  gausses,  when  the  magnetizing  force  is  known 
in  gilberts  per  centimeter.  Such  curve  sheets  will  be  found 
in  handbooks,  and  will  also  be  supplied  by  steel  manufac- 
turers in  order  to  give  accurate  data  concerning  the  elec- 
trical steel  which  they  manufacture.  For  practical  use, 
such  curves  are  often  plotted  in  other  units,  for  instance,  the 
flux  density  in  lines  per  square  inch  against  the  magnetizing 
force  in  ampere-turns  per  inch.  The  flux  density  in  lines 
per  square  inch  is  equal  to  6.45  times  the  density  in  gausses. 
The  gilberts  per  centimeter  magnetizing  force  is  QAir  times 
the  ampere-turns  per  centimeter,  or  0.47T/2.54  times  the 
ampere-turns  per  inch. 

Prob.  14-6.  Find  from  the  curves  the  permeability  of  sheet 
steel  when  the  magnetizing  force  is  5,  10  and  15  ampere-turns 
per  inch.  What  is  the  magnetizing  force  in  gilberts  per  centi- 
meter in  each  case? 

Prob.  15-6.  What  magnetizing  force  is  necessary  to  produce 
a  flux  density  of  108,000  lines  per  square  inch  in  soft  cast  steel? 
A  density  of  100,000  lines  per  square  inch?  Repeat  for  66,000 
and  58,000  lines  per  square  inch.  What  is  the  flux  density  in 
gausses  in  each  case? 

64.  Ampere-turns  to  Produce  a  Given  Flux.  The  fact 
of  the  dependence  of  the  permeability  of  iron  upon  the 
flux  density  often  introduces  a  difficulty  into  magnetic- 
circuit  calculations.  We  cannot  apply  Ohm's  law  for  the 


156      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

magnetic  circuit  unless  we  know  the  permeability  of  the 
material.  We  cannot  tell  what  the  permeability  will  be 
unless  we  know  the  flux  density.  One  sort  of  problem  is 
quite  simple.  For  instance,  if  the  total  flux  is  given,  we 
can  compute  the  flux  density,  look  up  the  corresponding 
value  of  permeability  for  the  material  used  on  the  curve 
sheet,  and  then  use  Ohm's  law  for  computing  the  required 
magnetomotive  force  and  hence  the  number  of  turns  and 
the  current  necessary  in  a  magnetizing  coil  in  order  to  pro- 
duce this  flux. 

A  simpler  proceeding  is  to  make  use  of  the  B-H  curves 
and  determine  directly  the  ampere-turns  per  centimeter 
(H)  necessary  to  produce  the  desired  flux  density  (B). 
By  multiplying  this  value  (H)  by  the  length  of  the  mag- 
netic circuit,  the  total  number  of  ampere-turns  (NI)  is 
obtained. 


FIG.  63.     A  current  in  the  coil  will  set  up  a  magnetic  flux  in 
the  iron  circuit. 

Referring  to  Fig.  63,  suppose  it  is  required  to  find  the 
current  necessary  in  the  magnetizing  coil  of  200  turns  in 
order  to  produce  a  total  flux  in  the  circuit  of  60,000  lines. 
The  core  we  will  assume  to  be  similar  to  that  used  in  a  trans- 
former and  built  up  of  steel  laminations.  The  cross-sec- 
tional area  of  the  core  is  six  square  centimeters,  and  the 


THE  MAGNETIC  CIRCUIT  157 

desired  flux  density  is  hence  10,000  gausses.  Referring  to 
the  curve  for  sheet  steel  in  Fig.  62,  we  find  the  corresponding 
permeability  to  be  4500.  The  mean  length  of  the  magnetic 
circuit  is  approximately  40  centimeters.  Therefore  the 
reluctance  is 

40 
91  -  '  a°0148  °ereted 


and  the  necessary  magnetomotive  force  is 

y  =  &0  =  0.00148  X  60,000  =  88.8  gilberts. 

The  current  flowing  in  the  magnetizing  coil  must  therefore 
be 

trr  QQ  o 

7  =  "I  03^200  =  °'353  ampere' 


The  same  result  may  be  obtained  more  directly  as  follows. 
From  the  B-H  curve  for  sheet  steel,  Fig.  61,  we  find  that 
for  a  flux  density  (B)  of  10,000  gausses  there  are  required 
4.50  ampere-turns  per  inch  of  magnetic  circuit.  If  the 
length  of  the  path  is  40  centimeters,  the  total  number  of 
ampere-turns  is 

40 
NI  =  ^-rj  X  4.50  =70.8  ampere-turns. 

^.O'x 

The  necessary  current  is  then 
70.8 


200 


=  0.354  ampere. 


Prob.  16-6.  Referring  to  Fig.  63,  what  current  must  flow 
in  the  coil  if  the  flux  is  to  be  50  %  greater  than  the  value  used 
in  the  above  example? 

Prob.  17-6.  A  certain  magnetic  circuit  is  made  up  of  an- 
nealed-steel  sheets.  The  length  of  the  circuit  is  2  feet  and  the 
cross-section  is  5  square  inches.  How  many  ampere-turns  are 
necessary  in  order  to  set  up  a  flux  of  500,000  maxwells  in  this 
circuit? 


158      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  18-6.  (a)  If  the  cross-section  of  the  magnetic  path  in 
Prob.  17-6  were  twice  as  great,  how  many  ampere-turns  would 
be  needed  to  set  up  twice  as  many  maxwells?  (6)  How  do  you 
account  for  your  answer,  inasmuch  as  there  is  twice  as  much 
iron  being  magnetized  to  the  same  flux  density  as  in  Prob.  17? 

Prob.  19-6.  The  U-shaped  magnet  of  Fig.  64  must  have 
100,000  maxwells  in  the  air  gaps  at  ends  A  and  B.  The  core 
C  of  sheet  steel  has  a  mean  length  of  4  inches  and  cross-section 


FIG.  64.     A  U-shaped  electromagnet. 

of  1  square  inch.  The  pieces  D  and  E  are  of  cast  steel,  each  hav- 
ing a  length  of  3  inches  and  a  cross-section  of  1  \  square  inches. 
The  length  of  the  air  gaps  at  A  and  B  is  0.04  inch  each,  and  the 
area  of  each  gap  may  be  taken  as  1 J  square  inches.  The  length 
of  the  magnetic  path  through  the  cast-steel  piece  F  is  4  inches 
and  the  mean  cross-section  is  If  inches.  How  many  turns 
must  be  wound  on  each  of  the  cores  D  and  E  if  the  coils  are  to 
carry  0.45  ampere? 

65.  Flux  Produced  by  a  Given  Number  of  Ampere-Turns. 

The  reverse  problem  to  those  in  the  above  article  is  also 
very  simple  when  the  magnetic  circuit  is  uniform,  as  is  the 
case,  for  instance,  in  the  transformer  core  shown  in  Fig. 
63.  Suppose,  for  example,  that  it  is  required  to  find  the 
flux  which  will  be  produced  in  this  core  when  the  magnet- 
izing coil  carries  a  current  of  1  ampere.  With  this  current 


THE  MAGNETIC  CIRCUIT 


159 


there  are  200  ampere-turns  in  the  magnetizing  coil,  which 
accordingly  has  a  magnetomotive  force  of 

0.47T  200  =  251  gilberts. 
The  magnetizing  force  is  hence 

251 

-^-  =  6.27  gilberts  per  centimeter. 

Referring  to  Fig.  61,  this  is  found  to  correspond  to  a  flux 
density  of  13,300  gausses.  The  total  flux  produced  in  the 
core  will  thus  be  13,300  times  6  or  79,800  maxwells. 

When,  however,  the  magnetic  circuit  is  not  uniform,  it  is 
necessary  to  resort  to  a  cut-and-try  process  of  solution  in 
order  to  find  the  amount  of  flux  which  will  be  produced  by  a 


2cm. 


FIG.  65.      The  magnetic  circuit  has  not  a  uniform  cross-section. 

given  magnetomotive  force.  It  will  be  best  to  illustrate 
what  is  meant  by  a  problem.  In  Fig.  65  is  shown  again  a 
simple  transformer  core,  but  in  this  case  the  cross-section 
of  the  core  is  not  uniform.  It  is  required  to  find  the  flux 
density  produced  by  a  magnetizing  coil  as  shown.  The 
magnetomotive  force  is  evidently 

0.47T  20X15  =  377  gilberts. 
The  magnetic  circuit  can  be  divided  into  two  parts,  the  first 


160      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

being  approximately  30  centimeters  long  and  of  4  square 
centimeters  cross-section,  the  second  14  centimeters  long 
and  of  8  square  centimeters  cross-section.  In  estimating 
these  lengths,  it  is  necessary  to  allow  for  a  certain  amount 
of  rounding  of  flux  at  the  corners  of  the  core.  The  amount 
to  be  allowed  in  each  case  depends  to  a  considerable  extent 
upon  what  experience  has  shown  to  be  correct  for  similar 
problems.  A  fairly  accurate  estimate  can  usually  be  made 
by  laying  out  the  core  to  scale  and  drawing  in  the  probable 
flux  paths. 

Since  there  are  two  parts  to  the  circuit  arid  we  do  not 
know  the  permeability,  it  is  now  necessary  to  start  at  the 
other  end  and  work  back.  Let  us  assume  as  a  first  guess 
that  there  is  a  total  flux  of  60,000  lines.  Referring  to  our 
curve  sheet,  and  multiplying  out,  we  find  that  this  cor- 
responds to  a  magnetomotive  force  of  390  gilberts  for  the 
first  part  of  the  circuit  and  of  14  gilberts  for  the  second 
part;  that  is,  a  total  magnetomotive  force  of  404  gilberts. 

This  first  guess  was  evidently  too  high,  for  we  have  377 
gilberts  available.  It  is  thus  necessary  to  decrease  our  esti- 
mate of  the  flux  lines.  We  should  naturally  make  the 
next  guess  by  decreasing  the  first  guess  approximately  by 
a  percentage  of  difference  between  the  computed  and  avail- 
able gilberts.  In  this  manner  we  obtain  as  a  second  guess 
59,000  lines  for  the  flux.  Computing  as  before  we  find  that 
this  corresponds  to  374  gilberts.  Since  this  is  a  close 
enough  check  to  our  available  gilberts  for  ordinary  purposes, 
we  may  consider  the  problem  solved  and  the  second  estimate 
for  flux  as  correct. 

By  using  the  above  method  of  calculation,  and  bearing 
in  mind  always  the  analogous  solution  of  an  electric  circuit, 
it  is  possible  to  solve  even  rather  complicated  magnetic 
circuits. 

A  type  of  problem  whicn  often  comes  up  is  such  as  is 
illustrated  in  Fig.  66,  where  the  iron  circuit  is  the  core  of  a 
three-phase  transformer.  Suppose  it  is  required  to  find 


THE  MAGNETIC  CIRCUIT 


161 


the  total  flux  in  leg  A  of  this  core,  due  to  a  known  magneto- 
motive force  applied  by  a  coil  as  shown.  It  is  necessary  to 
use  the  cut-and-try  process  which  may  be  outlined  as  follows . 
Assume  a  certain  flux  density  in  leg  C.  By  the  use  of 
curves,  as  in  the  preceding  problem,  compute  the  magneto- 


FIG.  66.     The  magnetic  circuit  of  a  three-phase  transformer. 

motive  force  necessary  to  force  this  flux  density  through  a 
length  of  circuit  shown  by  the  dotted  line  in  the  figure. 
This  same  magnetomotive  force  may  then  be  considered  to 
be  applied  to  core  B,  for  as  far  as  the  magnetic  circuit  is 
concerned,  the  legs  B  and  C  are  parallel  paths  for  the  flux, 
with  the  same  magnetomotive  force  applied  to  each.  Di- 
viding this  magnetomotive  force  by  the  length  of  the  core  B, 
we  find  the  magnetizing  force  on  B  in  gilberts  per  centi- 
meter. The  curves,  Fig.  61,  will  then  give  us  the  flux  den- 
sity in  B,  and  multiplying  by  the  cross-sectional  area  we 
get  the  total  flux  in  leg  B.  Adding  the  flux  in  legs  B  and 
C  we  have  the  flux  in  A.  This  gives  us  the  flux  density  in 
A,  and  computing  as  before,  we  find  the  magnetomotive 
force  for  the  part  of  the  path  shown  by  the  dot-dash  line. 
Adding  together  these  two  magnetomotive  forces,  we  have 
the  total  magnetomotive  force.  If  this  corresponds  with 
the  applied  magnetomotive  force,  then  our  estimate  of  the 
flux  density  was  correct;  otherwise  the  assumed  flux  den- 


162      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

sity  should  be  corrected  in  proportion  to  the  error  found  in 
the  computed  magnetomotive  force. 

It  will  often  be  found  in  solving  problems  of  this  sort  that 
an  allowance  may  be  made  for  the  curvature  of  the  mag- 
netization curves  which  are  being  used,  and  the  work  thus 
shortened  by  rendering  successive  estimates  more  accurate. 
This  should  be  clear  from  Fig.  67,  for  if  we  find  that  <f>c  is 
too  great  and  4>B  too  small,  and  we  take  a  mean,  OE,  of 
the  magnetomotive  force,  the  flux  will  be  greater  than  our 
expectation  by  FG. 


O  A  E  D 

FIG.  67.     The  magnetization  curve  of  iron. 


The  work  may  also  be  shortened  by  making  the  first 
guess  by  neglecting  the  effect  of  the  center  leg  and  consider- 
ing the  simple  remaining  circuit.  The  resulting  flux  den- 
sity gives  a  point  to  start  from  in  the  cut-and-try  process. 

Prob.  20-6.  In  Fig.  66  the  magnetizing  coil  is  of  200  turns 
and  carries  a  current  of  2  amperes.  The  material  of  the  core 
is  annealed  sheet  steel.  The  paths  A  and  C  are  each  50  cm. 
long,  and  leg  B  is  20  cm.  long.  The  cross-section  of  the  mag- 
netic circuit  is  uniform  throughout,  and  of  16  sq.  cm.  area. 
What  will  be  the  flux  density  in  each  part  of  the  circuit?  (If 
the  computations  check  the  given  magnetomotive  force  within 
4%,  the  results  are  as  accurate  as  can  be  expected  io  practice 
in  view  of  probable  variations  in  the  quality  of  the  steel.) 


THE  MAGNETIC  CIRCUIT 


163 


66.  Air  Gaps.  In  the  above  consideration,  the  magnetic 
circuit  was  entirely  in  iron.  It  is  very  common,  as  we  know, 
for  air  gaps  to  be  interposed  in  an  iron  magnetic  circuit. 


FIG. 


Frame  of  a  two-pole  motor  showing  field  coils  and  poles. 


Such,  for  instance,  is  the  case  in  the  magnetic  circuit  of  the 
dynamo  shown  in  Fig.  68.  The  magnetic  circuit  of  this 
machine  is  indicated  in  Fig.  69  by  the  dotted  lines,  and 


FIG.  69.     The  magnetic  paths  of  the  motor  of  Fig.  68. 

it  will  be  noted  that  it  is  made  up  of  the  following  parts: 
an  armature  core,  an  air  gap,  a  pole  shoe,  a  pole  core,  two 
half  yokes  in  parallel,  a  second  pole  core,  a  second  pole 
shoe  and  a  second  air  gap,  all  of  these  parts  being  in  series. 
To  compute  the  magnetomotive  force  necessary  to  pro- 


164      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

duce  a  given  flux  in  such  a  magnetic  circuit,  we  compute 
the  magnetomotive  force  necessary  for  each  part,  and  add. 
In  computing  the  air  gap,  we  have  noted  above  that  the 
permeability  is  unity.  The  formula  for  the  magnetomotive 
force  necessary  for  the  air  gap  is  thus  simply 

3=  Bl  gilberts,  (13) 

and  for  the  ampere-turns  for  the  air  gap 

Ni  =  -QJ-  Bl  ampere-turns,  (14) 

where  B  is  the  flux  density  in  gausses  and  I  is  the  length 
in  centimeters. 

The  correct  cross-section  to  use  for  an  air  gap  in  com- 
putation work,  however,  is  sometimes  a  matter  of  doubt. 
This  is  for  the  reason  that  the  flux  in  the  air  gap  spreads 
out.  It  is  exactly  as  if  we  were  studying  an  electric  current 
in  an  electric  circuit  composed  of  carbon  with  a  small  air 
gap  in  the  carbon  conductor,  the  whole  being  immersed  in  a 
salt  solution.  Between  the  ends  of  the  carbon  rod,  the 
current  would  spread  out  somewhat.  Similarly,  in  an  air 
gap  in  an  iron  magnetic  circuit,  the  flux  spreads  out  as 
shown  in  Fig.  70.  Where  the  air  gap  is  short,  it  is  usually 
sufficiently  accurate  to  increase  each  dimension  used  in  the 
computation  by  the  length  of  the  air  gap.  For  Fig.  70, 
the  air  gap  would  hence  be  considered  as  of  length  5,  and 
cross-sectional  area 

A  =  (b  +  5)  (a  +  5). 

When  an  air  gap  is  introduced  into  a  magnetic  circuit, 
there  are  further  effects  noticeable  by  which  the  magnetic 
flux  immediately  makes  its  presence  felt.  Pieces  of  iron 
near  the  gap  are  attracted.  In  fact,  there  is  a  strong  at- 
traction between  the  two  parts  of  a  magnetic  circuit  sep- 
arated by  the  gap,  the  amount  of  which  we  will  shortly 
compute.  The  two  ends  of  the  iron  part  of  the  magnetic 


THE  MAGNETIC  CIRCUIT 


165 


circuit  act  like  the  ends  of  the  familiar  permanent  magnet, 
except  that  they  may  be  made  much  more  powerful.  One 
of  these  will  be  a  north  pole  and  the  other  a  south  pole. 


ir  Gap 


FIG.  70.     Note  that  the  flux  spreads  out  in  the  air  gap. 

In  order  to  distinguish,  it  is  convenient  to  draw  arrowheads 
on  the  flux  lines  in  such  a  manner  that  the  pole  at  which  the 
flux  lines  leave  the  iron  will  be  a  north  pole. 

It  will  now  be  found  that  there   is  a  right-hand- screw 
relation  between  the  direction  of  the  current  in  the  mag- 


FIG.  71. 


Showing  the  relation  between  the  direction  of  the  magnetic 
flux  and  the  electric  current  producing  it 


netizing  coil  and  the  direction  of  the  flux  produced  in  a 
magnetic  circuit.  This  relation  is  shown  in  Fig.  71.  It 
may  be  expressed  as  follows. 

The  flux  produced  by  a  current  in  a  magnetizing  coil  will 
bear  the  same  relation  to  the  current  that  the  motion  of  a 
nut  (Fig.  72)  on  a  right-hand  screw  bears  to  the  rotation  of 


166      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

the  nut.     If  the  nut  is  turned  in  the  direction  of  the  current 

in  the  coil,  its  progress  along  the  screw  will  be  in  the  direction 

of  the  flux  in  the  core. 

It  is,  of  course,  possible  for  a  coil  carrying  current  to  be 

threaded  by  flux  in  the  direction  reverse  to  the  above.     This 

may  be  accomplished  if  we  over- 
come the  magnetomotive  force  of 
the  coil  by  a  second  opposed  and 
more  powerful  magnetomotive  force. 
When  the  flux  is  produced  by  the 
magnetomotive  force  of  the  coil 
itself,  however,  it  will  have  the 

right-hand-screw    relation    to    the 
FIG.  72.    The  right-hand      ,r     ,.          .  ., 

direction  of  the  current    as   above 


outlined. 

As  has  been  noted,  the  permeability  of  iron  is  several 
hundreds  to  several  thousands,  while  the  permeability  of 
air  is  unity.  If  the  air  gap  is  any  large  part  of  a  magnetic 
circuit,  the  reluctance  of  the  air  gap  will  hence  constitute 
almost  the  entire  reluctance  of  the  entire  circuit.  Suppose 
that  we  have  a  magnetic  circuit  of  total  length  ten  inches 
with  an  air  gap  one-tenth  of  an  inch  long.  The  length  of 
the  air  gap  is  one  percent  of  the  length  of  tlie  magnetic 
circuit.  Yet  if  the  permeability  of  the  iron  is  a  thousand 
and  the  cross-section  of  the  magnetic  circuit  is  uniform 
throughout,  the  reluctance  of  the  air  gap  will  be  ten  times 
that  of  the  remainder  of  the  circuit. 

This  leads  to  a  convenient  simplification  in  the  cut-and-try 
process  for  solving  a  magnetic  circuit.  When  such  a  circuit 
contains  an  air  gap  of  fair  size,  so  that  its  reluctance  is  a 
considerable  proportion  of  the  reluctance  of  the  magnetic 
circuit,  a  first  estimate  of  the  flux  density  may  be  found  by 
considering  the  magnetomotive  force  to  be  applied  to  the 
air  gap  alone. 

Prob.  21-6.  In  the  magnetic  circuit  of  Fig.  63,  an  air  gap 
is  now  introduced  in  the  middle  of  one  of  the  long  legs  by  cut- 


THE  MAGNETIC  CIRCUIT  167 

ting  a  space  1  millimeter  wide  entirely  through  with  a  hacksaw. 
What  current  will  now  be  necessary  in  the  magnetizing  coil  to 
force  a  flux  of  60,000  lines  through  the  magnetic  circuit? 

Prob.  22-6.  If  the  gap  in  Prob.  21-6  were  only  TV  milli- 
meter wide,  what  current  would  be  necessary? 

Prob.  23-6.  If,  with  this  narrow  gap  of  Prob.  22-6,  a  cur- 
rent of  0.4  ampere  flows  in  the  coil,  what  total  flux  will  be  pro- 
duced? 

67.  Leakage  Flux.  In  an  electric  circuit,  practically 
none  of  the  electric  current  leaks  off  the  electric  circuit 
under  ordinary  conditions,  that  is,  with  a  metallic  conductor 
and  ordinary  insulating  materials.  The  conductivity  of 
the  metal  may  be  ten  million  mhos  per  inch  cube,  while  the 
conductivity  of  the  surrounding  insulating  material  is  one 
ten-millionth  of  a  mho  per  inch  cube.  These  values  are 
so  far  apart  that  at  ordinary  differences  of  potential  and 
ordinary  values  of  current  the  amount  of  leakage  is  entirely 
negligible.  The  current  is  the  same,  therefore,  in  all  parts 
of  the  circuit,  and  no  conduction  takes  place  outside  the 
circuit  except  when  the  potential  gradient  becomes  abnor- 
mally high. 

In  the  magnetic  circuit  an  entirely  different  condition 
prevails.  The  permeability  of  the  iron  used  for  the  mag- 
netic flux  path  may  be  several  thousands,  but  the  permea- 
bility of  the  surrounding  material  will  be  approximately 
unity.  The  leakage  paths  may  be  short  and  of  large  cross- 
section,  and  their  reluctance  hence  fairly  small.  There- 
fore in  all  ordinary  magnetic-circuit  calculation,  it  is  neces- 
sary to  take  account  of  the  amount  of  flux  which  passes  in 
leakage  paths.  The  total  flux  will  not  pass  completely 
around  the  circuit,  but  some  of  it  will  leak  out,  and  hence 
this  loss  of  flux  must  be  taken  into  account  in  computing 
the  flux  density  of  the  circuit  itself.  There  is  no  magnetic 
insulator.  By  this  we  mean  simply  that  there  is  no  ma- 
terial which  has  a  permeability  so  small  that  practically  no 
flux  can  be  passed  through  it.  All  materials  except  a  few 


168      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

magnetic  materials  such  as  cobalt,  nickel  and  iron,  have 
permeabilities  very  close  to  one.  Even  a  vacuum  has  a 
permeability  differing  from  that  of  air  by  an  amount  so 
small  that  the  difference  is  detected  with  difficulty. 

Magnetic  leakage  is,  of  course,  particularly  high  when 
conditions  in  the  circuit  are  such  as  to  produce  large  mag- 
netic differences  of  potential  between  portions  of  the  circuit 
in  close  proximity.  Such  large  differences  of  magnetic 
potential  may  be  produced  by  introducing  an  air  gap  into 
the  circuit,  or  by  an  arrangement  where  the  magnetomotive 
forces  of  two  coils  are  large  and  opposed  to  each  other. 

In  the  magnetic  circuit  of  a  dynamo,  as  shown  in  Fig. 
69,  the  principal  reluctance  of  the  magnetic  circuit  occurs 
at  the  air  gaps.  The  pole  shoes  are  hence  at  a  large  differ- 
ence of  magnetic  potential.  This  means  simply  that  there 
is  a  large  magnetomotive  force  acting  between  pole  shoes, 
and  for  this  reason  there  will  be  a  large  leakage  flux  between 
the  pole  shoes;  that  is,  considerable  flux  will  pass  from  shoe 
to  shoe  without  going  through  the  armature.  Only  that  por- 
tion of  the  flux  is  useful  which  passes  through  the  armature. 

The  leakage  flux  between  two  points  of  a  magnetic  circuit 
may  be  computed  by  finding  the  magnetic  difference  of 
potential  between  the  two  points,  and  also  the  reluctance 
of  the  leakage  path  joining  them.  The  magnetic  difference 
of  potential  between  two  points  of  a  magnetic  circuit  may 
be  found  in  exactly  the  way  that  we  find  the  electric 
difference  of  potential  between  two  points  on  an  electric 
circuit.  It  is  equal  to  the  reluctance  drop  in  the  circuit 
joining  it,  unless  there  is  a  magnetomotive  force  in  that 
part  of  the  circuit.  In  such  a  case,  the  magnetic  potential 
difference  is  the  magnetomotive  force  acting  taken  together 
with  the  reluctance  drop.  Kirchhoff  s  law  can  be  used  for 
the  magnetic  circuit. 

The  magnetic  potential  difference  between  the  points  A 
and  B,  Fig.  73,  can  thus  be  written  as 

0"  -  4>9UB,  (16) 


THE  MAGNETIC  CIRCUIT 


169 


or 


QAirNI  - 


(17) 


because  OAirNI  is  the  magnetomotive  force  applied  to  that 
part  of  the  circuit  and  <f>9lAB  is  the  reluctance  drop  of  that 
part  of  the  circuit.  The  expression  CF  —  4>9lAu)  is  thus  the 
rise  in  potential  from  B  to  A.  If  the  points  A  and  B  are 
at  the  terminals  of  the  coil,  the  expression  £F  —  <^AB  might 
be  called  the  terminal  magnetic  potential  of  the  coil.  This 


FIG.  73.     The  flux  lines  <£3  and  fa  are  called  leakage  lines. 

is  analogous  to  the  terminal  electric  potential  of  a  generator, 
which  is  equal  to  the  e.m.f.  of  the  generator  minus  the  in- 
ternal resistance  drop  of  the  generator,  or  E  —  IR. 

The  magnetic  potential  difference  between  the  points 
A  and  B  can  also  be  computed  from  the  fall  of  magnetic 
potential  from  A  to  B  along  the  path  ADCB,  because  the 
fall  in  potential  from  A  to  B  must  equal  the  rise  in  potential 
from  B  to  A  as  computed  above. 

The  potential  difference  of  A  and  B  must  thus  equal 

(18) 


where  $1  =  flux  through  the  iron  from  A  to  D;  &AD 


170      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

reluctance  of  the  iron  from  A  to  D;  fa  =  flux  through  the 
armature;  Rarm  =  reluctance  of  the  armature,  etc. 

In  addition  to  the  quantities  of  flux  <£,  0,,  and  fa,  there 
will,  however,  be  other  quantities  of  flux  through  the  air 
between  certain  points  on  the  iron  part.  For  instance, 
the  flux,  04,  leaks  from  the  region  of  the  pole  core  to  the 
end,  B,  of  the  yoke,  because  there  is  a  difference  of  potential 
between  these  points  of  (AirNI  —  <f>\9lAB)  gilberts.  To 
determine  the  amount  of  this  flux,  04,  the  magnetic  potential 
difference  must  be  divided  by  the  reluctance  of  the  leakage 
path  from  A  to  B.  This  reluctance  is  very  difficult  to  com- 
pute. 

In  a  similar  manner,  some  flux,  fa,  leaks  between  the  pole 
tips  D  and  C  without  going  through  the  armature.  The 
amount  of  this  flux  is  equal  to  the  magnetic  potential  dif- 
ference of  the  pole  tips  (2fa9lgap  +  fa9^rm)  divided  by  the 
reluctance  of  the  leakage  path,  which  again  is  difficult  to 
determine  accurately. 

The  reluctance  of  the  leakage  path  in  air  between  two 
parts  of  a  magnetic  circuit  is  usually  difficult  to  compute. 
The  flux  in  the  air  spreads  out,  and  as  a  result  the  cross- 
section  of  the  gap  is  usually  a  matter  which  is  somewhat 
difficult  to  estimate.  In  a  few  cases  which  are  simple  in 
geometric  form,  the  value  of  the  reluctance  has  been  accu- 
rately computed.  In  most  cases  which  arise  in  practice, 
the  geometry  of  the  circuit  is  so  complicated  that  an  accu- 
rate solution  is  impossible  and  approximation  must  be  used. 

As  an  example  illustrating  this  proposition,  the  alternating- 
current  transformer  will  serve  excellently.  This  is  one  of 
the  most  important  pieces  of  electrical  apparatus  and  the 
flexibility  which  it  gives  to  the  commercial  circuit  is  the 
principal  reason  that  a  wide  use  of  alternating  currents 
prevails.  The  transformer  consists  of  two  and  sometimes 
more  coils  wound  on  a  single  magnetic  circuit.  One  of  these, 
the  primary,  supplies  a  current  which  magnetizes  the  core, 
and  since  this  current  is  alternating  or  varying,  the  flux  in 


THE  MAGNETIC  CIRCUIT 


171 


the  core  will  vary  with  it.  This  varying  flux  produces 
a  voltage  in  the  secondary  coil.  By  varying  the  number 
of  turns  on  the  secondary,  we  can  vary  the  voltage  pro- 
duced, and  so  obtain  any  desired  ratio  of  voltages  in  the 
unit.  When  a  current  is  drawn  from  the  secondary,  more 
current  will  flow  in  the  primary.  As  a  result,  the  two  coils 
may  each  be  carrying  a  heavy  current  at  the  same  time. 
Their  magnetomotive  forces  will  be  opposed  but  not  quite 
equal.  The  difference  between  them  is  the  magnetomotive 
force  which  forces  the  flux  about  the  circuit.  Such  an 
arrangement  is  shown  in  Fig.  74.  In  a  transformer  the 
leakage  flux  is  of  great  importance,  since  it  determines 


1.5  Amperes 


FIG.  74.  The  diagrammatic  representation  of  a  transformer. 
The  currents  in  the  two  coils  tend  to  set  up  opposing  fluxes 
in  the  core. 

almost  entirely  the  regulation  of  the  transformer.  In  prac- 
tice, the  coils  are  split  up,  and  part  of  the  primary  and  also 
part  of  the  secondary  put  on  each  leg  in  order  to  keep  the 
leakage  flux  small.  It  will  be  more  convenient,  however, 
for  us  to  examine  a  more  simple  arrangement  in  which  the 
primary  is  wound  on  one  leg  and  the  secondary  on  the  op- 
posite leg.  The  arrangement  shown  in  the  figure  will  hence 
have  a  much  larger  leakage  flux  than  that  which  would  be 
allowable  in  practice,  but  is  simpler  to  compute  than  the 
actual  arrangement. 

We  have  chosen  an  instant  when  the  primary  coil  of  500 
turns  is  carrying  a  current  of  20  amperes  and  the  secondary 
coil  of  5000  turns  is  carrying  1.5  amperes.  The  magneto- 
motive forces  will  be  in  the  directions  shown  by  the  dotted 


172      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

arrows,  and  are  therefore  opposed.  The  net  ampere-turns 
useful  in  forcing  the  flux  about  the  circuit  will  be 

(20  X  500)  -  (1.5  X  5000)  =  2500  ampere-turns, 

The  magnetic  circuit,  allowing  for  curvature  at  the  corners, 
is  about  350  centimeters  long.  The  magnetizing  force  is 
therefore 

2500 


~-  —  - 

oOU 


8.98  gilberts  per  centimeter. 


Neglect  for  a  moment  the  variation  in  density  of  the  flux  at 
different  parts  of  the  core  due  to  leakage  and  refer  to  the 
curves  for  sheet  steel,  and  it  is  found  that  this  magnetizing 
force  corresponds  to  a  flux  density  of  14,500  gausses,  which 
is  about  the  maximum  usually  used  in  transformer  practice. 
The  cross-section  of  the  iron  is  300  square  centimeters. 
Thus  there  are  4,350,000  maxwells  total  flux. 

Let  us  now  examine  the  magnetomotive  force  acting  on  the 
air  path  from  point  A  to  point  B  on  the  core.  The  left- 
hand  coil  supplies  10,000  ampere-turns.  About  1250  of 
these  are  used  up  in  forcing  the  flux  through  the  reluctance 
of  half  of  the  electric  circuit  as  shown  by  the  dotted  lines; 
that  is,  the  reluctance  drop  around  this  half  of  the  circuit  is 
1250  ampere-turns  or  1570  gilberts.  The  net  ampere-turns 
back  from  A  to  B  is  hence 

10,000  -  1250  =  8750  ampere-turns.  (33) 

This  value  might  also  have  been  found  by  taking  the  mag- 
netomotive force  of  the  secondary,  7500  ampere-turns,  and 
adding  to  it  the  reluctance  drop  of  1250  ampere-turns  in 
the  right-hand  half  of  the  core.  The  reluctance  drop  in 
this  case  is  added  to  the  magnetomotive  force  of  the  coil 
since  the  flux  is  being  forced  through  the  coil  against  its 
magnetomotive  force.  This  is  entirely  analogous  to  the 
potential  drop  caused  in  a  battery  in  an  electric  circuit  in 
which  the  current  is  being  forced  through  the  battery  in  a 


THE  MAGNETIC  CIRCUIT 


173 


direction  opposite  to  its  generated  potential.     The  magneto- 
motive force  between  points  C  and  D  will  similarly  be  the 
magnetomotive  force  in  the  primary  minus  the  reluctance 
drop  in  the  left-hand  leg.     The 
magnetomotive  force  across  the 
air  path  is  thus  seen  to  vary 
linearly    along    the    core,    and 
since     the     flux     produced     is 
directly    proportional     to     the 
magnetomotive  force,  it  will  be 
accurate  for  us  to  use  the  mean 
value,  that  is,  the  8750  ampere- 
turns  acting  at  the  center,  as 
shown  in  Fig.  75. 

We    now   need    to    estimate 
the  reluctance   of  the   leakage 

path.     In  Fig.  76  is  shown  a  cross-section  of  the  trans- 
former  with   the   flux   lines    drawn  in   their   approximate 


Distance  along  Core 

FIG.  75.  Showing  the  values 
of  the  magnetomotive  forces 
between  different  points  on 
the  core  in  Fig.  74. 


FIG.  76.     Showing  the  flux  distribution  in  the  leakage  paths  of  the 
transformer  of  Fig.  75. 

positions  for  the  leakage  flux.     The  value  to  be  chosen  for 
the  cross-section,  that  is,  the  amount  to  be  allowed  for 


174      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

fringing,  is  largely  a  matter  of  experience,  and  we  will 
consider  it  more  fully  below.  At  present  we  will  take  the 
value  of  30  centimeters  as  a  reasonable  width,  and  90  cen- 
timeters as  a  reasonable  length  of  the  cross-section  of  the 
leakage  path.  As  the  average  length  of  the  leakage  path 
we  might  take  65  centimeters.  Accordingly,  the  reluctance 
of  this  leakage  path  is 

;       a  -  h  =  i  x  36o5x  90  =  °-024  oereted> 

and  the  total  leakage  flux  is 


The  flux  in  the  core  was  estimated  at  4,350,000  maxwells. 
The  leakage  flux  is  thus 

458  000 

'  nnrt  or  about  10  percent  of  the  main  flux. 


In  an  accurate  computation  it  would  now  be  necessary  to 
recalculate  the  main  flux,  for  since  the  flux  is  not  exactly 
uniform  throughout  the  core,  the  assumptions  we  made  in 
our  first  calculation  were  not  entirely  accurate.  The  correc- 
tion will  be  small,  however,  and  we  will  not  estimate  it  at 
this  time. 

Prob.  24-6.  In  the  transformer  shown  in  Fig.  74,  the  sec- 
ondary current  is  increased  from  1.5  to  1.75  amperes,  the  pri- 
mary current  remaining  as  before.  What  will  be  the  new 
values  for  the  main  and  leakage  fluxes? 

Prob.  25-6.  Fig.  77  represents  a  magnetic  path  in  a  4-pole 
generator.  On  each  pole  are  wound  770  turns  carrying  5 
amperes.  The  reluctance  drops  for  various  parts  of  the  mag- 
netic circuit  are  as  follows: 

Each  air  gap,  ..............  2582  ampere-turns; 

Armature,  ................   298  ampere-turns  ; 

Each  pole,  ................   775  ampere-turns. 

(a)  What  is  the  reluctance  drop  for  the  yoke?  (6)  What  is  the 
magnetic  potential  between  tips  of  adjacent  poles? 


THE  MAGNETIC  CIRCUIT 


175 


Prob.  26-6.     The  average  distance,  I,  between  nearest  edges 
of  adjacent  poles  of  the  generator  in  Fig.  77  is  3.28  inches. 


FIG.  77.     The  magnetic  paths  in  a  four-pole  generator. 


The  average  cross-section  of  the  path  of  the  leakage  flux  from 
tip  to  tip  is  7  X  0.5  inches.  How  great  is  the  leakage  flux 
from  tip  to  tip? 


SUMMARY    OF   CHAPTER  VI 

AN  ELECTRIC  CIRCUIT  ALWAYS  LINKS  ONE  OR  MORE 
MAGNETIC  CIRCUITS  and  sets  up  magnetic  fluxes. 

NO  ENERGY  IS  CONSUMED  by  a  magnetic  circuit  in 
which  the  flux  is  not  changing. 

A  CHANGE  IN  THE  FLUX  in  a  magnetic  circuit  induces 
an  electromotive  force  in  any  interlinking  circuit. 

WHEN  A  CHANGE  IN  THE  FLUX  INDUCES  ONE  AB- 
VOLT  in  an  interlinking  electric  circuit  of  one  turn,  the  rate 
of  change  is  said  to  be  ONE  MAXWELL  PER  SECOND. 

THE  MAXWELL,  sometimes  called  a  LINE,  is  the  unit  of 
magnetic  flux.  The  letter  <£  is  used  to  denote  magnetic  flux. 

THE  GAUSS  is  the  unit  of  flux  density  and  equals  one 
maxwell  per  square  centimeter.  The  letter  B  is  used  to  denote 
flux  density. 

MAGNETIC  FLUX  IS  SET  UP  BY  MAGNETOMOTIVE 
FORCE  much  as  an  electric  current  is  set  up  by  an  electromo- 
tive force. 

OHM'S  LAW  FOR  THE  MAGNETIC  CIRCUIT: 

Magnetomotive  force 

Magnetic  flux  =  ^  ,      > 

Reluctance 


-a' 

where       0  =  the  magnetic  flux  in  MAXWELLS, 

y  =  the  magnetomotive  force  in  GILBERTS, 
91  =  the  reluctance  in  OERSTEDS. 

The  quantity  0&  is  called  the  RELUCTANCE  DROP  and 
is  analogous  to  IR,  the  resistance  drop.* 

THE  MAGNETOMOTIVE  FORCE  can  be  found  from  the 
following  equation : 

4.NI 

10 
176 


THE  MAGNETIC  CIRCUIT  177 

where         fF  =  the  magnetomotive  force  in  GILBERTS, 

N  =  the    number   of   turns   around   the    magnetic 

circuit, 

I  =  the   electric   current   flowing  in  the   turns   in 
AMPERES. 

THE  RECIPROCAL  OF  THE  RELUCTANCE  OF  A  UNIT 
CUBE  OF  A  MATERIAL  IS  CALLED  ITS  PERMEABILITY. 
THE  RELUCTANCE  CAN  BE  FOUND  from  the  equation 


where       91  =  the  reluctance  in  OERSTEDS, 

I  =  the  length  of  the  magnetic    path  in  CENTI- 

METERS, 

H  =  the  permeability  of  the  substance  in  GAUS- 
SES DIVIDED  BY  GILBERTS  PER  CENTI- 
METER, 

A  =  the  cross-section  area  in  SQUARE  CENTI- 
METERS. 

SERIES  AND  PARALLEL  ARRANGEMENTS  OF  RE- 
LUCTANCES are  treated  in  a  manner  similar  to  that  used  for 
series  and  parallel  arrangements  of  resistances. 

THE  MAGNETOMOTIVE  FORCE  PER  CENTIMETER 
of  the  magnetic  circuit  is  called  THE  MAGNETIZING  FORCE 
and  is  represented  by  the  letter  H. 

THE  PERMEABILITY  may  be  thought  of  as  B/H.  For 
all  non-magnetic  substances  (including  a  vacuum)  it  has  sub- 
stantially the  value  of  unity.  For  magnetic  substances  it 
varies  from  unity  to  several  thousand  depending  upon  the 
nature  of  the  substance  and  the  degree  of  magnetization. 

MAGNETIZATION  CURVES  plotted  between  simultane- 
ous values  of  B  and  H  are  of  great  value  in  determining  the 
ampere-turns  necessary  to  produce  a  given  flux  in  a  given 
magnetic  circuit. 

IN  MAGNETIC  CIRCUITS  CONTAINING  AIR  GAPS, 
the  greater  part  of  the  magnetomotive  force  is  usually  con- 
sumed by  the  gap. 

THERE  IS  NO  MAGNETIC  INSULATION.  Thus  there 
is  always  a  certain  amount  of  FLUX  LEAKAGE  from  the 
magnetic  path  in  which  it  is  desired  to  confine  the  flux, 


178      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

LEAKAGE  FLUX  can  be  computed  by  dividing  the  magneto- 
motive force  across  the  leakage  path  by  the  reluctance  of  the 
leakage  path.  This  reluctance  is  usually  difficult  to  determine 
and  is  generally  estimated. 


PROBLEMS   ON   CHAPTER  VI 


Prob.  27-6.  A  D'Arsonval  galvanometer,  such  as  is  used 
in  the  laboratory,  has  been  calibrated  and  it  is  found  that  one 
centimeter  deflection  corresponds  to  2.1  X  103  abvolt-seconds. 
A  measuring  coil  of  25  turns  is  placed  on  an  iron  core  and  the 
ends  of  the  coil  are  connected  to  the  ballistic  galvanometer. 
The  core  is  now  suddenly  magnetized,  and  a  deflection  of  20.2 
centimeters  is  obtained.  The  core  area  is  0.8  square  inch. 
What  is  the  flux  density  produced  in  the  core? 

Prob.  28-6.  A  telephone  transformer  is  in  common  use 
with  the  following  combinations  of  turns: 


Primary  Taps 

Secondary  Taps 

No. 

Turns 

No. 

Turns 

1-2 
2-3 
3-4 

150 
300 
600 

5-6 

6-7 

7-8 

1200 
2400 
4800 

If  the  flux  in  the  core  were  changing  at  a  uniform  rate,  with  a 
voltage  of  1.2  appearing  on  taps  1  and  2t  what  would  be  the 
voltage  measured  on  the  various  pairs  of  secondary  taps? 

Prob.  29-6.  Determine  the  ampere-turns  necessary  to  pro- 
duce a  flux  density  of  70  kilolines  to  the  square  inch  in  the 
core  A  of  the  cast-steel  chuck  of  Fig.  148.  (Mean  lengths  and 
cross-sections  of  flux  paths  must  be  estimated.) 

Prob.  30-6.  Find  the  number  of  ampere-turns  necessary 
to  produce  a  flux  density  of  65  kilolines  in  the  central  core 
of  the  annular  chuck  of  Fig.  149  if  a  disk  of  machined  steel  one 
inch  thick  and  of  the  same  diameter  as  the  chuck  is  laid  on  the 
face  of  the  chuck  and  in  intimate  contact  with  it. 


Prob.  31-6.     Work  Prob.  20-6  using  3000  ampere-turns, 
the  cut-and-try  method. 

179 


Use 


180      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  32-6.  Cut  a  f  inch  gap  in  leg  C  of  Fig.  66  and  re- 
calculate Prob.  31-6. 

Prob.  33-6.  A  transformer  has  a  continuous  winding  of 
700  turns.  At  every  one  hundred  turns  there  is  taken  out  a 
tap  or  lead.  Numbering  these  leads  1,  2,  3,  4,  5,  6,  7,  8,  if  the 
flux  is  changing  at  such  a  rate  that  a  voltage  of  110  is  read  be- 
tween leads  1  and  7,  what  voltages  will  be  read  at  the  same 
instant  between  leads  1  and  2,  6  and  7,  2  and  5? 

Prob.  34-6.  What  is  the  magnetic  potential  difference  be- 
tween the  centers  of  the  pole  cores  in  Prob.  25-6? 

Prob.  35-6  Fig.  78  and  79  represents  a  simple  2-pole  mag- 
netic chuck,  frame  and  cover.  The  pole  pieces  are  of  cast  steel, 
the  "  work  "  is  of  wrought  iron.  The  length  of  each  core  is  2f 
inches,  the  cross-section  of  each  core  is  2  x  4j  inches.  The 
cover  and  frame  are  f  inch  thick.  The  space  between  poles 
is  1  inch  wide.  The  "  work  "  is  4f  X  5  X  li  inches.  How 
many  ampere-turns  must  be  wound  on  each  core  in  order  to 
set  up  a  flux  of  540,000  maxwells  through  a  cross-section  at 
the  center  of  the  "  work"?  Assume  no  leakage  and  no  air 
gaps  between  "  work "  and  cover  or  between  cover  and 
cores. 

Prob.  36-6.  From  the  data  in  Prob.  35-6,  what  magnetic 
difference  of  potential  exists  between  (a)  points  6  and  c  of  the 
chuck  in  Fig.  78?  (6)  Points  a  and  d?  (c)  Points  g  and  h?  (<Q 
Points  k  and  11 

Prob.  37-6.  Compute  the  leakage  between  core  and  shell 
in  Fig.  78,  disregarding  that  which  leaks  between  "  work  " 
and  shell. 

Prob.  38-6.     Compute  the  leakage  between  cores  in  Fig.  78. 

Prob.  39-6.  Leakage  flux  leaves  and  enters  magnetic  ma- 
terial nearly  at  right  angles  to  the  surface.  This  causes  the 
leakage  lines  of  flux  between  the  "  work  "  and  cover  to  have 
the  shapes  shown  in  Fig.  78.  The  lines  may  be  assumed  straight 
in  the  space  between  fa  and  ed,  and  arcs  of  concentric  circles 
with  a  as  a  center  between  fa  and  am.  Compute  the  reluctance 
of  the  path  thus  formed  between  the  4j  X  li  sides  of  the 
"  work  "  and  the  cover  of  the  chuck.  (Note:  integrate  be- 
tween limits  0  and  arc  mf.) 

Prob.  40-6.  Compute  the  leakage  from  the  4j  X  1^-inch 
sides  of  the  "  work  "  to  the  cover  of  the  chuck  in  Fig.  78. 


THE  MAGNETIC  CIRCUIT 

e         Work  Non-Magnetic 


Pole 


181 

Cover  and 


Pole  Piece 


FIG.  78.     Vertical  section  of  a  magnetic  chuck  and  "work." 


r 


J 


FIG.    79.     Top  view  of  magnetic  chuck  of  Fig.  78. 


182      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  41-6.  Compute  the  leakage  from  the  5  X  Ij-inch 
sides  of  the  "  work  "  to  the  cover  in  Fig.  78. 

Prob.  42-6.  Taking  into  account  the  leakage  flux  as  found 
in  Probs.  37-6,  38-6,  40-6  and  41-6,  make  a  closer  approxi- 
mation of  the  ampere-turns  necessary  to  set  up  540,000  lines 
through  a  cross-section  at  the  center  of  the  "  work  "  than  was 
possible  in  Prob.  35-6,  where  the  leakage  flux  was  neglected. 


0.6cm. 


FIG.  79a.    A  magnetic  circuit  made  up  of  two  parallel  paths. 

Prob.  43-6.  By  what  percentage  will  the  reluctance  of  the  ring 
of  Fig.  79a  be  changed  if  the  material  of  higher  permeability  is 
placed  on  the  inside?  The  volume  of  each  material  and  the  over- 
all dimensions  of  the  ring  remain  the  same. 


CHAPTER  VII 
THE   MAGNETIC   FIELD 

In  this  chapter  will  be  discussed  methods  for  determining 
the  flux  density  at  various  points  hi  magnetic  fields  produced 
by  electric  currents. 

68.  The  Line  Integral  Law.  Kirchhoff's  law  applied  to 
magnetic  circuits  states  that  the  sum  of  the  reluctance  drops 
around  a  closed  magnetic  circuit  is  equal  to  the  sum  of  the 
magnetomotive  forces  acting  on  the  circuit.  If  the  magnetic 
circuit  is  uniform,  this  is  simply 

F=£0,  (1) 

which  is  another  way  of  writing  the  familiar  expression 

OAwNI 
*=     "§T 

This  expression,  we  have  seen,  is  true  whether  the  magnetic 
circuit  is  uniform  or  not.  In  the  case  of  a  non-uniform 
magnetic  circuit,  we  take  the  several  parts  which  are  in 
series  and  multiply  the  flux  by  the  reluctance  of  each  and 
add,  in  order  to  obtain  the  total  magnetomotive  force  ac- 
curately; that  is, 

(3) 


In  the  case  of  a  magnetic  circuit  varying  uniformly  in  cross- 
section,  such  as  a  circuit  made  of  a  conical  piece  of  iron, 
where  the  reluctance  varies  continuously  from  point  to 
point,  the  law  becomes 


(4) 

183 


184      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Inserting  for  <f>  its  value 

<f>  =  J3A  f 
and  for  91 

and  canceling,  we  obtain 

y=/|<&  (5) 

Then  since  B  =  pH 

this  may  be  written 

/• 

(6) 


This  expression  may  be  stated  in  words:  the  line  integral 
of  H  about  a  closed  circuit  is  equal  to  fF;  or,  the  line  integral 
of  the  magnetizing  force  about  a  closed  magnetic  circuit  is 
equal  to  0.47T  times  the  number  of  ampere-turns  linking  the 
circuit. 

In  the  above,  we  have  assumed  that  H  is  always  in  the 
direction  of  the  length  of  the  magnetic  circuit.  The  law 
above,  however,  has  been  found  to  be  true  in  a  more  general 
way  than  this.  In  fact,  if  we  take  into  account  the  direc- 
tion of  H,  that  is,  consider  the  above  integral  to  be  a  line 
integral,  the  general  proposition  follows  that  the  line  in- 
tegral of  H  about  any  closed  curve  is  equal  to  OAw  times 
the  number  of  ampere-turns  linking  the  curve.  It  will  now 
be  necessary  to  explain  carefully  and  in  detail  just  what  is 
meant  by  this  expression,  for  it  is  important. 

First,  consider  what  is  meant  by  a  line  integral.  If  a 
point,  as  it  moves  along  a  path,  is  acted  upon  by  a  variable 
force,  the  line  integral  of  this  force  along  the  path  is  equal 
to  the  length  of  the  path  times  the  average  value  of  the 
component  of  the  force  in  the  direction  of  the  path.  In 
other  words,  if  at  every  point  of  the  path  we  take  the  force 
multiplied  by  the  cosine  of  the  angle  between  the  force  and 
the  direction  of  the  path,  and  integrate  this  product  over 


THE  MAGNETIC  FIELD  185 

the  entire  distance,  then  the  result  is  the  line  integral  of 
the  force  along  the  path.  This  can  be  made  clear  by  an 
example. 

Suppose  that  we  attach  a  rubber  band  to  a  fixed  circular 
post  as  shown  in  Fig.  80,  and  fasten  the  other  end  to  a 
pencil  point.  The  rubber  band  may  be  wrapped  around 


FIG.  80.  The  pencil  is  attached  by  a  stretched  rubber  band  to  the 
post.  The  line  integral  of  the  force  exerted  by  the  rubber  band  as 
the  pencil  moves  from  B  to  A  is  the  work  done  as  the  pencil  moves 
from  B  to  A. 

the  post  and  fastened  at  some  convenient  point.  The  pull 
of  the  rubber  band  exerts  a  force  on  the  pencil  in  the  direc- 
tion shown  by  the  arrow.  The  line  integral  of  this  force 
along  the  path  AB  is  the  work  done  on'  the  pencil  point  as 
it  moves  from  B  to  A.  If  we  represent  the  force  by  F  and 
the  work  by  W,  this  becomes 

W  =      B  Fdl,  (7) 


which  is  to  be  understood  as  a  line  integral  equation.  The 
line  over  the  F  indicates  that  F  is  a  vector  and  is  to  be  con- 
sidered as  such;  that  is,  it  has  direction  as  well  as  magnitude. 
This  line  integral  may  also  be  written 


W 


=    fB  FcoaOdl.  (8) 

JA. 


186      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

This  means  simply  that  the  path  is  divided  into  elementary 
lengths,  dl,  the  magnitude  of  the  force  at  each  one  of  these 
points  is  multiplied  by  the  cosine  of  the  angle  which  it  makes 
with  the  direction  of  the  path,  and  the  limit  of  the  sum  of 
all  these  products  is  taken.  The  line  integral  above  from 
A  to  B  is  evidently  the  work  done,  that  is,  the  energy  given 
up  by  the  rubber  band  as  the  pencil  point  passes  from  A  to  B. 
Suppose  now  that  the  pencil  point  is  moved  around  a 
closed  path  such  as  CDE.  The  line  integral  around  this 
closed  path  must  be  zero;  that  is, 

Fdl  =  0.  (9) 

CDEC 

We  know  that  this  must  be  true,  for  the  work  done  by  the 
rubber  band  is  evidently  zero,  since  it  returns  to  exactly  its 
original  condition.  To  state  the  matter  in  another  way, 
if  the  above  integral  were  not  zero,  that  is,  if  there  were  a 
total  amount  of  energy  produced  by  the  rubber  band  during 
the  operation,  then  it  would  be  possible  to  construct  a  per- 
petual motion  machine  by  arranging  the  point  to  continu- 
ally trace  this  closed  circuit,  and  such  a  machine  we  know  to 
be  impossible.  We  thus  know  that  the  line  integral  of  a 
force  such  as  the  above  around  such  a  closed  path  is  zero. 
Consider  now  the  integral  around  the  closed  path  FGH, 
which  links  the  post  to  which  the  rubber  band  is  attached. 
This  integral  is  not  zero,  for  when  the  point  has  returned  to 
point  F,  the  rubber  band  has  been  stretched  by  an  additional 
amount,  since  it  has  been  wound  one  more  turn  around  the 
post.  The  amount  that  it  has  been  wound  up,  that  is, 
2irr  times  the  force  necessary  to  stretch  it  a  unit  length,  is 
the  total  amount  of  work  done.  The  line  integral  is  equal 
to  this  work,  and  if  we  represent  by  K  the  force  necessary 
to  stretch  the  rubber  band  one  unit  of  length,  we  have  the 
expression 

"M   =   2irrK.  (10) 

FGHF 


THE  MAGNETIC  FIELD  187 

The  law  for  the  magnetic  field  is  very  similar  to  this. 
H  is  a  force,  called  the  magnetizing  force.  The  line  in- 
tegral of  this  force  about  a  closed  path  which  links  no  coils 
carrying  current  is  zero;  that  is, 

Hdl  =  0.  (11) 

The  line  integral  of  H  about  a  closed  path  linking  a  coil 
carrying  current  is  equal  to  the  magnetomotive  force  in  the 
coil,  that  is,  4irNI.  This  is  expressed  by 

Hdl  =  lirNL  (12) 

The  magnetomotive  force  F  must  therefore  be  work  rather 
than  force.  Similarly,  we  have  found  that  the  electromotive 
force  E  is  really  work  and  not  a  force  in  the  usual  meaning 
of  the  term. 


FIG.  81.  The  line  integral  along  any  path  in  the  magnetic  field  is  the 
work  done  by  the  magnetizing  force,  H,  acting  upon  a  particle  moved 
along  that  path. 

This  law  will  enable  us  to  examine  the  flux  distribution 
in  a  magnetic  circuit  of  very  irregular  shape,  in  particular 
for  a  magnetic  circuit  in  air,  where  the  lines  spread  out. 
When  a  coil  of  wire  carrying  current  is  simply  suspended  in 


188      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

the  air,  with  no  iron  present,  the  field  spreads  out  and  ex- 
tends a  considerable  distance  from  the  coil,  taking  somewhat 
the  shape  shown  in  Fig.  81.  We  may  compare  this  with  the 
electric  circuit  in  the  following  manner.  Suppose  that  an 
ordinary  flashlight  battery  of  cylindrical  form  is  dropped 
into  a  bucket  of  salt  water.  The  battery  will  be  short- 
circuited  by  the  salt  solution,  and  lines  of  current  flow  will 
extend  from  one  end  of  the  battery  to  the  other,  spreading 
out  through  the  solution  much  in  the  same  way  as  above. 
It  is  difficult,  and  in  fact  almost  impossible,  to  figure  the 
reluctance  of  the  path  for  a  solenoid  such  as  is  shown  in 
Fig.  81.  We  know,  however,  that  the  line  integral  of  H 
along  such  a  path  as  DCE  is  zero.  We  know  also  that  the 
line  integral  of  H  along  such  a  path  as  FGK  which  links 
the  coil  is  equal  to  QAw  times  the  product  of  the  current  in 
amperes  and  the  number  of  turns  in  the  coil.  By  utilizing  this 
law,  we  are  enabled  to  find  the  flux  density  in  the  vicinity  of 
wires  and  coils  carrying  current  when  there  is  no  iron  present. 

69.  Field  About  a  Long  Wire.  We  will  first  examine  the 
flux  density  in  the  vicinity  of  a  long  straight  wire  carrying 
current.  This  problem  is  important  in  the  examination  of 
transmission  lines,  telephone  lines  and  so  on.  We  have  all 
heard  the  sound  in  a  telephone  receiver  caused  by  the 
variation  in  the  current  taken  by  the  motors  of  a  trolley 
car.  This  sound  is  produced  by  the  fact  that  the  flux  sur- 
rounding the  trolley  wire  induces  a  voltage  in  nearby  tele- 
phone lines  and  causes  a  hum.  The  regulation  of  a  trans- 
mission line,  that  is,  the  drop  in  voltage  from  one  end  to 
the  other,  on  alternating  currents,  is  largely  dependent  on 
the  magnetic  field  surrounding  the  wire.  For  these  and 
similar  reasons  it  is  important  that  we  have  an  accurate 
formula  for  this  problem. 

When  a  wire  carrying  current  is  situated  at  a  distance 
from  magnetic  materials  and  from  other  wires,  the  magnetic 
flux  about  the  wire  is  arranged  in  circles  concentric  with  the 
wire.  This  must  be  true  from  symmetry,  but  we  can  check 


THE  MAGNETIC  FIELD 


189 


it  with  the  arrangement  shown  in  Fig.  82. 
passed  through  a  hole  in  a  card. 
The  latter  is  sprinkled  with  iron 
filings,  and   the  wire  caused  to 
carry  a  current.     When  the  card 

is  tapped,  it  is  found  that  the  

filings  arrange  themselves  in 
circles  as  shown,  proving  the 
field  to  be  circular.  In  order  for 
a  current  to  flow,  there  must,  of 
course,  be  a  return  wire  some- 
where to  complete  the  circuit, 
but  we  will  consider  this  return 
wire  to  be  so  far  distant  that 
its  magnetic  effect  may  be 
neglected.  The  whole  electric 
circuit  constitutes  one  turn. 
An  application  to  this  turn 
of  the  right-hand 


A  long  wire  is 


FIG.  82.  Showing  the  rela- 
tion between  the  direction 
of  the  current  in  a  wire  and 
that  of  the  magnetic  field 
set  up  by  the  current. 


FIG.    83. 


rule  of  the  previous 
chapter  shows  that  the  flux  lines  must  have 
the  direction  shown  by  the  arrows.  This 
rule  may  now  be  stated  another  way.  The 
flux  about  a  long  wire  has  a  direction  the 
same  as  the  rotation  of  the  head  of  a  right- 
handed  screw  when  turning  in  such  a  direc- 
tion as  to  progress  in  the  direction  of  the 
current  along  the  wire. 

We  will  now  examine  the  strength  of  this 

circular  field.     In  Fig.  83,  we  draw  a  circle 

The  of  radius  r  about  the  wire.     The  magnetizing 


magnetizing  force  H  will  be  the  same  at  each  point  on 
force  at  a  dis-  ^^  circle,  since  the  flux  lies  in  concentric 
circles  about  the  wire,  and  it  will  be  every- 
where tangent  to  this  circle.     Hence  the  line 
integral  for  H  about  this  circle  is  simply  equal 
to  H  times  the  circumference  of  the  circle.    This  line  integral 


a  current  /. 


190        PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

must  be  equal  to  QAirNI,  and  since  the  entire  electric 
circuit  makes  only  one  turn,  N  may  be  taken  as  unity. 
We  have  therefore 

CHdl  =  QATrNI,  (13) 

or 

27rrH  =  0.47T/,  (14) 

from  which  it  follows  that 

H-?£  (15) 

where  /  is  in  amperes.  If  I  is  in  abamperes,  the  formula 
becomes 

•   --,  H=~r-  '  (16) 

If  the  wire  is  in  air,  where  the  permeability  is  everywhere 
unity,  B  would  always  be  equal  numerically  to  H,  as  given  by 


that  is,  the  flux  density  at  any  point  near  a  long  wire  in  air 
carrying  current  is  equal  to  0.2  times  the  current  in  amperes, 
divided  by  the  distance  from  the  wire  in  centimeters. 

If  the  wire  is  surrounded  by  a  material  of  a  permeability 
other  than  unity,  we  have 

*-|.57.  (18) 


Such  a  case  arises  when  a  conductor  is  at  the  center  of  a 
cable  with  an  iron  sheath  as  shown  in  cross-section  in  Fig. 
84.  If  the  thickness  of  the  sheath  is  a,  the  total  flux  per 
centimeter  of  length  in  the  sheath  will  be 

(19) 

27 
It  will  be  instructive  to  derive  the  formula,  H  =  —  > 


THE  MAGNETIC  FIELD 


191 


in  a  slightly  different  way.     Referring  to  Fig.  85,  the  mag- 
netizing force  at  the  point  P  at  the  distance  r  from  a  long 


Iron  Sheath 


FIG.  84.     A  copper  conductor  within  an  iron  sheath. 

wire   carrying  current  /  may  be  found  by  integrating  the 
effect  of  small  elements  of  the  wire  of  length  dx,  provided 


FIG.  85. 


To  find  the  magnetizing  force  at  P  produced  by  current  / 
in  conductor  X. 


we  make  an  assumption  as  to  the  effect  of  each  of  these 
elements.  Such  an  assumption  is  that  the  effect  of  an  ele- 
ment of  length  dx  carrying  a  current  /  at  a  point  distant  I 


192      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

is  equal  to  the  product  Idx  divided  by  I2  and  multiplied  by 
the  cosine  of  the  angle  0  which  a  perpendicular  to  the  wire 
makes  with  the  length  I 

(20) 


Under  this  assumption,  the  effect  of  the  total  wire  will  be 
found  by  adding  the  effects  of  all  the  elements,  that  is, 
integrating  for  x  along  the  total  length  of  the  wire.  This 
gives 


From  the  geometry  of  the  figure  we  have 

x  =  r  tan  0,  (22) 

and  differentiating 

dx  =  r  sec20  d0.  (23) 

Also 

cos  0 
Inserting  these  values  in  the  above  integral,  we  have 

f*  —  T  pn<3  f)  T  ctpp'ty? 

U    -        I     9          l  JQ  O^ 

H  =    I     2 — d0,  (^5; 


2 

where,  since  the  variable  has  been  changed  to  angle  0,  the 
limits  of  the  integral  are  the  limits  of  this  angle,  or  from 
minus  to  plus  a  right  angle.  This  integral  simplified  be- 
comes 


-  d0,  (26) 

t/_l         T 

2 

which,  integrating,  gives 


*  Note  from  Fig.  85  that  the  quantity  dx  cos  0  is  the  projection  of 
dx  upon  a  line  perpendicular  to  I. 


THE  MAGNETIC  FIELD 
Inserting  the  values  of  the  limits,  we  have 


193 


(28) 


which  is  the  same  result  as  obtained  previously  by  a  different 
method.  Our  expression  for  the  effect  of  an  element  of 
circuit  is  therefore  justified,  namely,  that  the  magnetizing 
force  varies  inversely  as  the  square  of 
the  distance  from  the  element;  that  it 
is  proportional  to  the  cosine  of  the  angle 
of  projection  of  the  element  and  propor- 
tional to  the  length  of  the  element  and 
to  the  current  carried.  We  may  there- 
fore use  the  expression  with  assurance 
in  subsequent  calculations. 

This  expression  for  the  magnetizing 
force  near  a  long  wire  carrying  current 

holds  strictly  only  for 

a  wire  of  infinite  length. 

For  ordinary  accuracy 


FIG. 


The  con- 
ductor subtends  an 
angle  of  2a  at  P. 


FIG.  87.  The  con- 
ductor subtends 
an  angle  at  P  of 


angle  of  2a,  we 

to  +  a,  and  we  have 


it  holds  for  a  wire  which  is  fairly  long  com- 
pared with  the  distance  from  the  wire  con- 
sidered. For  example,  if  we  are  examining 
the  field  strength  ten  centimeters  from  a 
wire  which  is  ten  meters  long,  the  formula 
will  hold  to  much  greater  than  engineering 
accuracy.  We  may,  however,  obtain  the 
effect  of  a  short  wire.  This  may  be  done 
by  simply  changing  the  limits  on  the 
integral  above.  Thus,  referring  to  Fig.  86, 
if  the  wire  subtends  at  the  point  P  a  total 
put  the  limits  of  integration  from  —  a 


H 


f*-\-ot  T 

Llr 


cos  B  det 


(29) 


194      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


which  integrated  gives 


rr      27  . 

H  =  —  sin  a. 


(30) 


This  formula  may  be  used  for  the  magnetic  effect  of  a 
straight  wire  of  finite  length.  If  the  point  P  is  not  situated 
opposite  the  midpoint  of  the  wire,  as  for  instance  in  Fig. 
87,  the  formula  evidently  becomes 


H  =  -  (sin 


sn 


(31) 


The  objection  may  properly  be  raised  that  it  is  not  cor- 
rect to  speak  of  the  magnetic  effect  of  a  short  straight  wire 
alone  since  electric  current  must  always  flow  in  a  closed 


FIG.  88. 


To  find  flux  density  at  P,  the  center  of  a  square  of  wire  carry- 
ing a  current. 


circuit.  A  short  straight  wire  may,  however,  become  a 
part  of  a  closed  magnetic  circuit,  and  the  total  effect  of 
the  whole  circuit  is  found  by  the  sum  of  the  effects  of  its 
parts.  Suppose,  for  example,  that  we  wish  to  find  the 
flux  density  at  the  center  of  a  closed  square  of  wire  carrying 
a  current  7,  as  shown  in  Fig.  88.  This  will  evidently  be 
four  times  the  effect  of  one  side,  or 

7  . 


B  =  H  = 


sin 


THE  MAGNETIC  FIELD  195 

If  the  length  of  the  side  of  the  square  is  a,  we  shall  have 


or  since  a  is  45°  when  point  P  is  at  the  middle  of  the  square, 
the  resulting  expression  for  the  flux  density  at  the  middle 
point  of  such  a  square  will  be 


when  the  square  of  wire  is  not  in  the  vicinity  of  iron.  In 
this  problem  the  effect  of  the  leads  by  which  the  current 
enters  the  square  cancel,  since  they  are  close  together  and 
the  current  flowing  in  them  is  in  opposite  directions. 


K— «— H 

FIG.  89.     A  and  B  represent  a  pair  of  parallel  conductors  carrying 
current.     The  current  goes  out  on  A  and  returns  on  B. 


It  is  now  easy  to  compute  the  total  effect  due  to  a  pair 
of  wires  such  as  are  shown  in  Fig.  89,  where  the  current  goes 
out  along  one  and  returns  by  the  other.  This  is  a  common 
arrangement  in  the  transmission  of  power,  and  the  consid- 
eration of  the  field  is  therefore  of  much  importance.  We 
will  assume  that  the  wires  are  separated  by  a  distance  D, 
their  radius  being  r.  The  flux  density  at  any  point  on  the 
line  joining  the  wires  at  a  distance  x  from  one  of  the  wires 
may  immediately  be  found.  The  magnetizing  effect  of  the 
left-hand  wire  we  have  just  found  to  be  equal  to 

HA=2-^-  (32) 

JU 

Since  the  wire  B  is  carrying  equal  current  in  the  opposite 
direction,  its  magnetizing  force  at  the  point  P  adds  directly 


196      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

to  that  of  A,  since  both  magnetizing  forces  are  perpendicular 
to  the  line  joining  the  centers  of  the  two  wires.  The  mag- 
netizing force  of  B  is. 

H*  =  W^x  (33) 

since  the  current  in  the  two  wires  is  7.  The  total  magnetizing 
force  is  hence 


Since  the  wires  are  in  air,  the  same  expression  holds  for  the 
flux  density;  that  is, 

B  =  2l(l  +  -i^—\  (35) 


The  total  flux  passing  between  the  two  wires  may  be  found 
by  integrating  this  expression,  thus 


/D~r  /1  1     \ 

G+^>        (36) 


-  loge(D  -  z)  (37) 


=  47  loge  —      —  maxwells  per  centimeter  of 
line, 

=  2  7  loge  --  -maxwells     per     centimeter 
length  of  wire.  (38) 

This  expression  will  be  of  use  to  us  later. 

The  field  at  some  point  not  on  the  line  joining  the  two 
wires  may  be  found  by  combining  the  values  of  H  at  the 
point  in  accordance  with  the  laws  of  vectors  as  shown  in 
Fig.  90.  Since  the  two  forces  now  are  not  in  the  same  direc- 
tion, they  must  be  added  vectorially  instead  of  algebraically. 


THE  MAGNETIC  FIELD 


197 


By  finding  the  value  of  H  and  then  the  value  of  B,  at  every 
point  P  in  the  vicinity  of  the  two  wires,  we  may  map  out 
the  entire  field.  This  distribution  of  the  flux  lines  about 
two  parallel  wires  carrying  current  in  opposite  directions 


FIG.  90. 


The  field  at  a  point  P  not  on  a  line  joining  A  and  B  is  the 
resultant  of  HB  and  H  A. 


may  thus  be  found  to  be  as  in  Fig.  91;  that  is,  a  series  of 
circles  with  their  centers  all  on  the  line  joining  the  centers 
of  the  wires  but  not  concentric  with  the  wires.  In  similar 


/ 

FIG.  91.     The  shape  of  the  magnetic  field  about  two  parallel  wires 
carrying  current  in  opposite  directions. 

manner  we  may  find  the  distribution  about  a  pair  of  wires 
carrying  current  in  the  same  direction,  and  it  will  be  found 
to  be  much  as  is  shown  in  Fig.  92.  We  know  that  two  paral- 
lel wires  carrying  current  in  opposite  directions  repel  each 
other,  and  that  two  parallel  wires  carrying  current  in  the 
same  direction  tend  to  be  drawn  together.  By  referring 


198      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

to  the  figures  showing  the  distribution  of  flux  about  such 
wires,  it  is  seen  that  this  action  may  be  interpreted  by  stating 
that  flux  lines  tend  to  crowd  apart  when  running  parallel  to 
each  other,  and  on  the  other  hand  tend  to  shorten  in  length. 
The  first  effect  tends  to  push  the  coils  of  wires  apart  when 
they  carry  current  in  opposite  directions,  and  the  second 


FIG.  92.     The  shape  of  the  magnetic  field  about  two  parallel  conductors 
carrying  current  in  the  same  direction. 

effect  tends  to  draw  wires  together  when  they  carry  current 
in  the  same  direction,  much  as  if  they  were  surrounded 
by  a  series  of  rubber  bands.  We  shall  later  compute  the 
amount  of  this  force  and  show  its  great  importance  in  elec- 
trical work. 

Prob.  1-7.  What  is  the  line  integral  of  H  at  a  radius  of  2.5 
inches  about  a  straight  wire  carrying  25  abamperes? 

Prob.  2-7.  In  order  to  produce  a  flux  density  B  =  6 
gausses  at  a  point  3  inches  from  the  center  of  a  straight  wire 
in  air  in  a  plane  perpendicular  to  the  wire,  what  must  be  the 
current  in  amperes? 

Prob.  3-7.  Plot  the  flux  density  for  points  on  a  line  per- 
pendicular to  the  axis  of  a  wire  which  is  carrying  a  current  of 
1  abampere. 

Prob.  4-7.  A  rectangle  3  by  5  inches  carries  a  current  of 
3  amperes.  What  is  the  flux  density  at  the  intersection  of 
the  diagonals  if  the  effect  of  the  leads  may  be  neglected? 

Prob.  5-7.  What  flux  is  enclosed  by  a  pair  of  No.  000  cop- 
per wires  spaced  6  feet  between  centers  and  of  one  mile  length 
each  when  /  =  30  amperes?  Neglect  the  end  effects. 


THE  MAGNETIC  FIELD 


199 


Prob.  6-7.  In  Fig.  93,  A  and  B  represent  the  "line  and 
return  "  of  a  single-phase  power  line  carrying  a  maximum  of 
48  amperes,  x  and  y  are  the  "  line  and  return  "  of  a  telephone 
line  running  parallel  to  the  power  line.  How  much  flux  links 
the  telephone  line  per  mile  when  the  maximum  current  is 
flowing  in  the  power  line? 


x    y 


FIG.  93.     A  and  B  represent  the  line  and  return  of  a  power  line,  x  and 
y  a  pair  of  telephone  wires. 

Prob.  7-7.  The  current  in  the  power  line  of  Prob.  6-7 
reverses  120  times  per  second  from  a  maximum  current  of  48 
amperes  in  one  direction  to  the  same  value  in  the  opposite 
direction.  What  average  voltage  is  induced  in  the  telephone 
circuit  if  it  is  20  miles  in  length? 

Prob.  8-7.  If  the  telephone  wires  of  Prob.  6-7  were  placed 
in  a  vertical  plane  40  feet  from  B,  with  x  at  a  distance  of  9  inches 
above  the  level  of  the  power  line  and  y  at  a  distance  of  9  inches 
below  it,  what  maximum  flux  per  mile  would  link  them? 

70.  Field  Inside  a  Conductor.  When  a  wire  carries  a 
current,  there  is  a  magnetic  field  produced  not  only  external 
to  the  wire  but  also  in  the  inte- 
rior of  the  wire.  We  will  now 
examine  the  strength  of  a  field 
on  the  inside  of  a  cylindrical  wire 
carrying  a  current. 

First,  let  us  consider  a  tubular 
conductor  such  as  is  shown  in 
Fig.  94.  We  will  assume  that 
the  current  density  is  uniform 
across  the  cross-section  of  the 
conductor.  For  any  circle  inside 
of  the  tube,  such  as  ABC,  the 
line  integral  of  H  will  be  zero,  for  this  circle  links  no  con- 
ductor carrying  current,  and  we  found  that  the  line  integral 


FIG.  94.     A  tubular  con- 
ductor. 


200      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

of  H  about  a  closed  path  is  always  equal  to  zero  when  the 
path  links  no  current.  There  is  no  current  linkage  in  this 
case,  for  all  the  current-carrying  material  is  external  to  the 
path  which  we  have  chosen.  Since  the  line  integral  of  H  is 
zero  about  this  circle,  it  follows  from  symmetry  that  H 
itself  must  be  zero  along  the  circle.  This  holds  for  any 
circle  which  is  entirely  inside  of  a  tube.  We  may  therefore 
conclude  that  the  magnetic  field  inside  of  a  tubular  con- 
ductor carrying  uniform  current  is  everywhere  zero. 

Consider  now  a  solid  cylindrical  con- 
ductor   carrying    a    current    uniformly 
distributed     across     its     cross-section. 
Such  a  conductor  of  radius  r  is  shown 
in  Fig.  95.     Let  us  examine   the   field 
strength  at  a  distance  a  from  the  center. 
Draw   a   circle   with   radius  a.      By 
FIG  QS^^Tsolid  con      symmetry  the  field  strength  is  the  same 
ductor  of  radius  r.      &t   every  point   on    this    circle    and   is 
everywhere  tangent  to  the  circle.     The 
line  integral  of  Ha  around  this  circle  is  accordingly  equal 
to  simply  Ha  times  the  circumference  of  the  circle;  that  is, 

=  2<n-aHa.  (39) 

We  know,  however,  that  this  line  integral  is  equal  to 
QAirNI,  where  N  is  the  number  of  turns,  (in  this  case  one), 
and  I  is  the  current  inside  of  the  path  of  integration.  The 
current  to  be  taken  in  this  case  is  hence  the  fraction  of  the 
total  current  which  lies  within  the  radius  a.  If  the  total 
current  in  the  wire  is  /  abamperes,  the  amount  within  the 
circle  of  radius  a  will  be 

a2 

-jj  /  abamperes.  (40) 


Therefore  we  have 

a5 


(41) 


J- 

•v 


or 


7  (  > 


THE  MAGNETIC  FIELD 


2Ia 


201 


(42) 


If  I  is  in  amperes  and  the  material  of  the  conductor  is 
copper  of  permeability  unity,  we  have 


D     0.21  a 

B  = gausses. 


(43) 


On  the  other  hand,  if  the  wire  is,  for  instance,  of  iron,  of 
permeability  other  than  unity,  we  must  take  the  permea- 
bility into  account  and  write 

0.2ju7  a 
r 


B  = 


(44) 


We  are  now  in  a  position  to  give  the  field  strength  at  any 
point  inside  or  outside  of  a  long  wire  carrying  current.  This 
may  be  shown  in  a  diagram,  as  in  Fig.  96,  where  the  flux 


FIG.  96.     The  flux  density  is  greatest  at  the  surface  of  the  wire  as  shown 
by  the  curve. 

densities  are  plotted  as  ordinates  and  the  distances  from 
the  center  of  the  wire  as  abscissas.  It  will  be  noted  that 
the  flux  density  follows  one  law  in  the  material  of  the  wire 
and  a  second  law  outside  of  a  wire,  so  that  there  is  a 


202      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


break  in  the  curve.  Note  also  that  when  x  =  r,  that  is,  at 
the  surface  of  the  wire,  the  results  obtained  from  these  two 
formulas  coincide  for  a  wire  of  unity  permeability. 

We  shall  see  later  that  the  distribution  of  flux  inside  a 
large  wire  is  an  important  factor  in  alternating-current 
work.  The  variation  of  this  flux  as  the  current  changes 
gives  rise  to  what  is  known  as  skin-effect.  By  this  effect, 
the  current  is  crowded  toward  the  surface  of  the  wire  and 
there  appears  a  much  larger  resistance  to  high-fre- 
quency alternating  currents  than  there  is  to  direct  cur- 
rents. 

Prob.  9-7.  In  the  wire  of  Prob.  3-7,  plot  the  flux  inside  the 
wire  with  the  center  as  an  origin,  magnifying  the  wire  so  as  to 
make  the  plot  reasonable  in  size. 

Prob.  10-7.  Derive  the  formula  for  the  total  flux  inside  a 
solid  round  wire  carrying  current. 

Prob.  11-7.     Derive  the  formula  for  the  total  flux  linkages 

inside  a  solid  round  wire  carrying  current. 

Mfe  -     T  >{  GL  J_    cJ 

71.  Flux  Density  at  the  Center  of  a  Single  Turn  of  Cir- 
cular Form  Carrying  Current.  It  will  now  be  necessary 

for  us  to  compute  the  flux 
density  existing  at  the  center  of 
a  single  turn  of  wire  of  circular 
shape  carrying  a  current  /,  as 
shown  in  Fig.  97.  The  other 
view  of  this  coil  is  shown  in 
Fig.  98,  with  the  direction  of 
the  flux  lines.  The  flux  at  P 

at   the   center   of   the   turn  is 
r  IG.  97.     A  single  circular  turn 

of  wire  carrying  a  current  /.       perpendicular   to   the   plane   of 

the  turn. 

The  element  of  length,  dl,  in  Fig.  97,  exerts  an  amount 
of  magnetizing  force  at  P  which  by  equation  (20)  is 


THE  MAGNETIC  FIELD  203 

Since  this  element  is  perpendicular  to  the  radius  r,  6  is  a 
right  angle  and  the  sine  of  6  is  unity,  this  becomes 

dl.  (45) 


FIG.  98.     A  side  view  of  the  wire  in  Fig.  97. 
The  total  effect  of  the  current  is  the  integral  of  d/f;that  is, 

//»2*r  T 
dH=Jg      i«B,  (46) 

where  the  integration  is  to  be  carried  once  around  the  circle. 
This  integrated  gives 

H=2-^-  (47) 

If  the  single  turn  is  in  air,  the  flux  density  is  also  given  by 
the  expression 

*-M,     .;  (48) 

where,  of  course,  /  is  in  abamperes.     If  /  is  in  amperes, 
this  becomes 

(49) 


204      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


1  cm. — >• 


Let  us  now  examine  the  magnetizing  force  produced  at  the 
center  of  an  arc  of  one  centimeter  radius  and  one  centimeter 
length,  as   shown   in  Fig.  99,  when  a 
current  of  one  abampere  flows.     Since 
the  leads  are  carried  off  on  radii  they 
will  exert  no  magnetizing  force  at  P. 
The  magnetizing  effect  of   the   arc    is 
found  by  an  integral  similar  to  that 
used  for  the   effect   of  a  single  turn, 
except  that  the  limits  of  integration  will 
FIG.  99.     Field  at  P     now  be  simply  zero  to  one;  that  is, 
is  produced  by  1  cm. 
arc  of  1  cm.  radius  TJ 

with  P  as  center. 


(50) 


which  gives  for  I  =  1,  r  =  1,  the  simple  expression 

H  =  1. 

The  effect  of  this  arc  of  a  circuit  is  therefore  to  produce 
a  unit  magnetizing  force  at  the 
center.  We  are  thus  led  to  a  new 
statement,  which  if  we  wish  may 
be  used  as  a  definition  of  unit  cur- 
rent: 

One  abampere  is  that  current  which 
flowing  in  a  unit  length  of  an  electric 
circuit  bent  into  an  arc  of  unit  radius 
will  produce  unit  magnetizing  force 
at  the  center  of  the  arc. 


The  flux  density  produced  in  air 
by  a  concentrated  coil  of  wire  as 
shown  in  Fig.  100  of  N  turns  and 
carrying  a  current  of  /  amperes,  is 
evidently 


FIG.  100.  A  circular 
concentrated  coil  of 
many  turns. 

(51) 


THE  MAGNETIC  FIELD 


205 


Prob.  12-7.  (a)  A  concentrated  coil  of  wire  having  36  turns 
and  carrying  a  current  of  0.3  ampere  produces  what  field  in- 
tensity at  the  center  of  the  coil?  (6)  What  flux  density? 
(c)  What  effect  would  a  different  permeability  of  the  medium 
surrounding  the  coil  have  on  flux  density?  Radius  is  r  cm. 

72.  Flux  Density  at  a  Point  on  the  Axis  of  a  Circular  Coil. 

We  have  found  the  flux  density  at  the  center  of  a  circular 
coil.  We  are  also  interested  in  the  flux  density  at  some  other 
point  on  the  axis  of  the  coil.  Such  a  coil,  of  a  single  turn,  is 
shown  in  Fig.  101  and  102,  and  we  will  proceed  to  compute 
the  flux  density  at  point  P  at  a  distance  of  a  centimeters 
from  the  center  of  the  coil. 


-a  dH 


FIG.  101.      A  single  circular 
turn  of  wire  carrying  current. 


FIG.  102.     Side  view  of  wire 
in  Fig.  101. 


Consider  an  element  of  the  coil  dl  and  call  6  its  distance 
from  P.  Call  0  the  angle  between  a  and  •&.  By  the  formula 
which  we  have  previously  used  for  the  magnetic  effect  of 
an  element  of  a  circuit  carrying  current,  we  may  write  the 
value  of  the  magnetizing  force  d  H  due  to  the  element  of  the 
circuit  dl.  This  will  be 


Idl. 


(52) 


that  is,  it  is  proportional  to  the  current  and  the  length  of 
the  segment,  and  inversely  proportional  to  the  square  of 
the  distance  from  the  point  to  the  element.  Since  6  is 
perpendicular  to  dl,  the  cosine  of  0,  which  previously  appeared 


206      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

in  the  formula  for  the  effect  of  an  element,  is  in  this  case 
unity. 

This  magnetizing  force  d  H  will  be  perpendicular  to  6.  It 
may  be  resolved  into  two  components,  d  Hv  perpendicular  to 
a  and  dHa  parallel  to  a.  From  the  geometry  of  the  fig- 
ure, the  vertical  component  will  be 


(53) 
and  the  horizontal  component 

dHa  =  dH  sine.  (54) 

The  total  magnetizing  force  at  P  will  be  the  vector  sum 
of  all  of  the  components  of  magnetizing  force  due  to  the 
element  dl  taken  completely  around  the  circle.  We  may 
note  that  to  every  element  dl  on  the  top  of  the  circle,  having 
a  vertical  component  of  magnetizing  force  dHv  directed 
upward,  there  will  be  an  opposite  element  on  the  other  side 
of  the  circle  which  will  have  an  equal  vertical  component 
of  magnetizing  force  directed  downward,  and  these  two 
components  will  cancel.  Hence  the  vertical  components  of 
magnetizing  force  for  the  whole  circle  cancel  out  completely. 
On  the  other  hand,  since  the  horizontal  components  are  all 
in  the  direction  of  the  axis  and  all  directed  toward  the  right, 
they  will  add  algebraically. 

In  order  to  find  the  total  magnetizing  force  at  P,  we  must 
therefore  add  the  effect  of  the  horizontal  components  due  to 
the  element  dl  taken  completely  around  the  circle;  that  is, 

TJ      C2irrjTT  -  r^IsmOjj  ,KK. 

H=  I      dHsm6=l      —  p—  dl.  (55) 

Integrating  this  expression  and  inserting  the  limits,  we  have 

/sin0 


(56) 

t/~ 

Since 

sin0=[,  (57) 


THE  MAGNETIC  FIELD  207 

we  have 

#  =  —  sin30;  (58) 

or  since  t 

6  =  Va2  +  r2,  (59) 

this  equation  becomes,  upon  substitution, 


which  is  our  final  formula  for  the  magnetizing  force  at  the 
point  P.  If  the  coil  is  surrounded  by  air,  the  flux  density 
will  also  be  given  by 

27rr2  1 
=  (a*  +  r*K 

If  for  the  single  turn  we  substitute  a  concentrated  coil  of 
N  turns,  the  total  flux  density  at  a  point  on  the  axis  will  be 
given  by 

2?r  r2  NT 


Prob.  13-7.  In  Fig.  101,  if  the  coil  has  5  concentrated  turns 
carrying  a  current  of  4  amperes,  what  is  the  flux  density  at  a 
point  P  where  a  =  6  inches  and  r  =  6  inches?  M  =  1. 

73.  The  Air-Core  Solenoid.  A  solenoid  is  a  coil  of  wire 
usually  wound  with  circular  cross-section  and  of  a  length 
great  in  comparison  with  its  diameter.  Solenoids  are  used 
on  relays,  circuit-breakers  and  similar  apparatus,  for  oper- 
ating the  mechanism  when  a  certain  amount  of  current 
flows  through  the  coil.  They  are  also  used  in  measuring 
instruments.  At  the  present  time,  we  will  confine  ourselves 
to  a  consideration  of  a  solenoid  with  an  air  core,  or  with  a 
core  made  of  non-magnetic  material.  Such  a  solenoid  is 
shown  in  Fig.  103. 

It  may  have  one  layer  or  several  layers.  For  simplicity 
we  will  consider  at  the  present  time  a  single-layer  solenoid. 
If  the  conducting  wire  were  rectangular  and  covered  with 


208      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

very  thin  insulation  so  that  a  cross-section  had  the  appear- 
ance of  Fig.  104,  the  current  around  the  coil  would  closely 


FIG.  103.     A  solenoid. 

approximate  a  "  current  sheet."     The  following  treatment 
assumes  such  a  current  sheet  to  flow,  although  most  coils  are 


FIG.  104.  A  solenoid  may  be  considered  to  be  made  up  of  square  wires 
having  a  very  narrow  space  between  them.  This  would  mean  that 
practically  a  sheet  of  current  flows  around  the  coil. 

constructed  of  round  wires  and  thus  depart  from  these  con- 
ditions. For  this  reason  the  following  treatment  is  an  approx- 
imation. 

< x — *l 


-GQOO 


oooloooooooooo 


l 


ooo 


oo 


oocpooooooo 

1 


FIG.  105.     Find  the  magnetizing  force  at  point  P,  the  center  of  the 

solenoid. 

We  will  first  investigate  the  field  strength  at  the  center 
point  of  the  solenoid.  Point  P,  Fig.  105,  is  the  central 
point  of  the  solenoid. 


THE  MAGNETIC  FIELD  209 

Consider  that  the  solenoid  is  I  centimeters  long  and  has 
N  turns,  which  carry  a  current  of  7  abamperes.  Consider 
a  belt  of  conductors  the  plane  of  which  is  at  a  distance  x 
from  point  P,  and  of  width  dx.  The  total  number  of  am- 
pere-turns comprised  in  the  belt  of  conductors  included  in 
the  length  dx  will  then  be 


. 

The  effect  of  this  belt  upon  the  point  P  is  the  effect  of  a 
circular  coil  upon  a  point  in  the  axis.  Call  this  effect  dH 
to  indicate  the  portion  or  the  magnetizing  force  at  P  which  is 
due  to  the  belt  of  conductors  within  the  region  dx.  Then 
from  the  formula  which  we  have  derived  for  such  a  circular 
coil,  we  shall  have 


dH=  sin3  (TT  -  ff)dx.  (63) 

We  may  reduce  this  to  an  expression  entirely  in  terms  of  9 
and  the  dimensions  of  the  coil,  as  follows.  From  the  triangle 
in  Fig.  105, 

x  =  r  cot  (TT  -  6)  =  -  r  cot  0,  (64) 

and  differentiating,  we  have 

dx  =  r  csc20  dB.  (65) 

Substituting  this  in  the  equation  above,  it  becomes 

ede.  (66) 


To  find  the  total  value  of  H,  we  integrate  this  expression 
to  cover  the  entire  coil.  If  8  increases  from  the  angle  a\  to 
angle  «2,  the  entire  coil  will  be  included.  This  leads  to  the 
following  integration 


I 


210      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


=  —  j  —  (cos  ai  —  cos  a2).  (67) 

This  resulting  expression  is  the  formula  for  the  magnetizing 
force,  H,  at  a  point  on  the  axis  of  a  long  solenoid.  If  the 
entire  surroundings  are  non-magnetic,  that  is,  if  there  is 
no  iron  present,  the  formula  gives  also  the  flux  density  at 
any  point  on  the  axis  of  a  solenoid  in  terms  of  the  angles 
«i  and  «2,  which  are  subtended  by  the  ends  of  the  coil. 
In  the  case  where  P  is  at  the  midpoint  of  the  coil, 

«2   =   7T   -  «i.  (68) 

From  this 

cos  «2  =  —  cos  ai  (69) 

and 

(cos  «i  —  cos  a2)  =  +  2  cos  «i.  (70) 

The  magnetizing  force  at  the   center  point  of  a  solenoid 
hence  becomes 

H=+^j£&»ai,  (71) 

or  in  terms  of  the  radius  and  length  of  the  coil 

(72) 

If  the  solenoid  is  very  long  in  proportion  to  its  diameter, 
say  twenty  times  its  diameter  in  length,  the  angle  a  is  very 
small  and  approximately 

cos  ai  =  1,  (73) 

which  gives 

(74) 
For  an  air-core  solenoid  in  which  no  magnetic  material  is 


THE  MAGNETIC  FIELD 


211 


present,  the  flux  density  will  be  equal  to  the  magnetizing 
force  and  we  may  write 


(75) 


This  formula  gives  the  flux  density  at  the  center  point  of  a 
long  solenoid  on  the  assumption  that  the  current  flows  in  a 
sheet  around  the  coil. 

We  will  now  examine  the  flux  density  at  some  other  point 
not  on  the  axis  but  in  the  cross-section  of  the  solenoid  at 
the  center.  In  Fig.  106  and  107,  P  is  such  a  point.  At  P 
pass  two  planes  each  parallel  to  the  axis  of  the  solenoid  and 
making  an  angle  of  d<}>  with  each  other,  as  shown  in  Fig. 
107.  These  will  cut  out  on  the  surface  of  the  solenoid  a 


ooo 


FIG.  106.  Find  the  flux  density  at  P,  a  point 
on  the  center  plane  but  not  on  the  long  axis 
of  the  coil. 


FIG.  107.  End  view 
of  coil  in  Fig. 
106. 


certain  strip  of  the  conductor  material  parallel  to  the  axis 
of  the  coil  and  of  width  dc. 

We  will  now  examine  the  portion  of  the  magnetizing  force 
at  P  produced  by  the  current  in  an  element  of  the  winding 
of  length  dc  and  width  dx,  as  shown  in  Fig.  107. 

The  magnetizing  force  due  to  an  element  we  have  pre- 
viously seen  is  proportional  to  the  current  in  the  element  and 
to  the  length  of  the  projection  of  the  element  perpendicular  to 
a  line  joining  it  with  the  point  in  question,  and  inversely 
proportional  to  the  square  of  its  distance  from  the  point. 
In  the  c.g.s.  system,  the  proportionality  factor  is  one. 


212      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

NI 

In  the  width  of  winding  dx  there  will  be  -y  dx  amperes. 

This  is  the  current  in  the  element.     The  projection  of  dc 
on  a  line  perpendicular  to  a  line  joining  it  with  P  is 

dc  cos  a, 

which  is  equal  to  ad(f>  since  d<f>  is  a  differential  angle.     The 
distance  from  the  element  to  P  is  equal  to  6. 

Accordingly  for  the  portion  of  the  magnetizing  force  due 
to  this  element  we  have 

dHp=^x^dx  (76) 

where  b2  =  (a?  -f  z2). 

This  magnetizing  force  will  not  be  parallel  to  the  axis  of 
the  solenoid,  but  will  have  a  component  which  is  parallel 
and  one  which  is  perpendicular  to  the  axis.  The  perpen- 
dicular component  need  not  be  considered  because  it  will 
cancel  out  when  the  entire  length  of  the  strip  on  the  surface 
of  the  solenoid  is  taken  into  account;  for  to  each  element 
dx  at  a  distance  x,  there  corresponds  an  element  on  the  op- 
posite side  of  the  center  and  at  an  equal  distance,  in  which 
we  have  the  magnetizing  force  directed  as  indicated  by  the 
dash-line  R  and  hence  with  an  equal  and  opposite  component 
perpendicular  to  the  axis.  We  need  therefore  consider  only 
the  horizontal  component.  To  obtain  the  component 
parallel  to  the  axis,  we  have  only  to  multiply  the  value  of 
the  force  d  HP  (Equation  76)  by  the  cosine  of  the  angle 
6,  Fig.  106.  Call  this  component  dH^.  Thus  the  mag- 
netizing force  parallel  to  the  axis  of  the  coil  at  any  point 
P  within  the  coil  and  due  to  current  in  an  element  of  the 
winding  is 

cos0dx.  (77) 


The  effect  of  a  strip  of  dc  width  and  running  the  entire  length 


THE  MAGNETIC  FIELD  213 

of  the  coil  can  be  found  by  integrating  this  expression  between 
the  limits  x  =  -  1/2  and  x  =  +  1/2. 

Thus 

Hd,=  C^x^cosedx.  (78) 

«y_L   '        o 

2 

This  integral  may  be  reduced  to  be  all  in  terms  of  0  by  making 
substitution  from 

x  =  a  tan  6,  (79) 

dx  =  a  sec2  6  d8, 
b  =  a  sec  0.  (80) 

Since  the  length  of  the  solenoid  is  great  compared  with 
the  diameter,  the  limit  of  the  angle  6  is  ir/2.  The  expression 
then  becomes 


t 

"2 

which,  integrated   in  exactly  the  same  manner  as  above, 
gives 

dH  =  — j-  d<f>.  (82) 

This  is  the  magnetizing  force  at  P  produced  by  that  por- 
tion of  the  conductor  which  lies  within  the  angle  d<f>.  In 
order  to  obtain  the  total  effect  of  the  solenoid,  we  need  now 
simply  to  integrate  0  from  0  to  2ir;  that  is, 


/2T 
«p 


d<f>  =  -^-P  -  (83) 


We  thus  find  that  the  magnetizing  force,  and  also  (in 
air)  the  flux  density,  at  the  point  P,  are  just  the  same  as  at 
a  point  in  the  center  of  the  solenoid.  It  follows  that  the 
flux  density  across  the  mid-section  of  a  long  solenoid  is 


214      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

uniform,  the  lines  are  parallel  to  the  axis  of  the  solenoid  at 
this  point,  and  the  total  flux  is 


(84) 


A 


We  are  led  to  the  remarkable  conclusion  that  a  long  solen- 
oid acts  as  if  it  had  a  reluctance  -j  •    The  flux  produced  at 

the  center  of  a  long  solenoid  is  exactly  the  same  as  would 
be  produced  if  all  of  the  lines  passed  completely  through  the 
length  of  the  solenoid,  thus  producing  uniform  flux  density, 
and  as  if  there  were  no  reluctance  whatever  in  the  path 
outside  the  solenoid  itself. 

We  know,  of  course,  that  the  path  outside  the  solenoid  has 
a  certain  small  reluctance.    It  is  true  also  that  the  lines  do 


FIG.  108.    Half  of  the  flux  which  passes  through  the  center  of  the  coil 
leaves  the  coil  before  reaching  the  ends  of  the  coil. 

not  all  completely  thread  a  solenoid,  but  half  of  them  leave 
through  the  sides  of  the  solenoid  as  shown  in  Fig.  108. 
It  happens,  however,  that  these  two  effects  exactly  offset 
each  other.  As  far  as  the  flux  at  the  mid-section  of  the 
solenoid  is  concerned,  we  may  therefore  treat  the  solenoid 
as  if  we  needed  to  consider  only  the  reluctance  of  the  air 
path  inside  of  the  winding. 

In  the  above  we  have  assumed  that  the  wires  were  small 
in  diameter  and  closely  spaced,  so  that  the  effect  was  to 
produce  practically  a  uniform  current  density  along  the 
surface  of  the  solenoid.  Our  formula  will  be  found  to  hold, 
therefore,  for  only  reasonable  distances  from  the  sides  of 
the  solenoid.  For  a  point  close  to  the  wires,  the  effect  of 


THE  MAGNETIC  FIELD  215 

the  size  of  the  wires  will  enter,  and  corrections  will  be  neces- 
sary to  the  formula  as  above  derived.  These  corrections 
are  too  involved  to  be  introduced  here.  A  careful  treatment 
of  this  matter  will  be  found  in  the  Bulletin  of  the  U.  S. 
Bureau  of  Standards  No.  1,  vol.  3,  page  3. 

Prob.  14-7.  Prove  the  statement  made  above  that  one- 
half  the  magnetic-flux  lines  leave  a  long  solenoid,  air-core  type, 
through  the  sides  of  the  coil. 

Prob.  15-7.  A  solenoid  has  a  length  of  11  inches  and  a 
radius  of  0.5  inch.  The  number  of  turns  to  the  inch  is  23. 
A  current  of  2.6  amperes  flows  through  the  coil.  What  is  the 
field  intensity  at  the  center  of  the  solenoid?  See  Fig.  106. 

Prob.  16-7.  What  is  the  total  flux  expressed  in  maxwells 
at  the  center  section  of  the  solenoid  of  Prob.  15-7? 

Prob.  17-7.  A  long  solenoid  is  desired  with  a  flux  at  the 
center  of  1000  lines.  The  diameter  must  be  1.7  inches  and  the 
number  of  turns  to  the  inch  25.  What  current  must  the  wire 
carry? 

74.  Calibration  of  a  Ballistic  Galvanometer.  The  above 
formula  for  the  flux  density  at  the  center  of  a  long  solenoid 
is  convenient  in  calibrating  a  ballistic  galvanometer  for  use 
in  measuring  flux.  It  was  noted  in  Art.  57  that  the  de- 

A 

oooooooooooooooooooooooooooooooooooooooooooooooooooooooooo 

00000000000000000000 

B 

ooooooooooooooooooo 
OOOOOOOOOOOOOOCX3OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO 

FIG.  109.     Coil  B  is  a  measuring  coil. 

flection  of  such  a  galvanometer  is  proportional  to  the  total 
change  of  flux  linkages  in  a  measuring  coil  connected  with  it. 
In  order  to  be  able  to  find  the  total  flux  by  this  method,  how- 
ever, it  is  necessary  to  calibrate  the  galvanometer,  that  is, 
find  its  deflection  for  some  known  change  of  linkages. 
For  this  purpose  we  may  use  an  arrangement  such  as  is 

NI 

shown  in  Fig.  109.  A  is  a  long  solenoid  with  -j-  ampere- 
turns  per  centimeter  of  length  when  connected  with  the 


216      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

battery  H  as  shown  in  Fig.  110.  B  is  a  measuring  coil  of  a 
number  of  turns  N'  and  cross-sectional  area  S,  placed  in- 
side and  close  to  the  center  of  A. 


FIG.  110.     Showing  how  coils  A  and  B  are  used  to  measure 
flux  in  ring  D. 

We  know  that  when  a  current  of  7  abamperes  is  flowing 
in  A,  the  flux  density  produced  at  the  center  will  be 

B  =  4w.  (85) 


The  total  flux  passing  through  B  is  accordingly 

*  -  4*^p<S,  (86) 

and  the  flux  linkages  thereby  produced 

T 


(87) 

In  practice,  the  coil  B  is  connected  in  series  with  the  gal- 
vanometer and  also  in  series  with  the  coil  D  enclosing  the 
flux  which  it  is  desired  to  measure. 

When  the  switch  is  closed,  forcing  a  current  of  /  abam- 
peres through  A,  there  will  be  a  deflection  of  G.  Call  this 
deflection  d.  By  a  similar  battery  arrangement,  the  flux 
may  now  be  caused  to  vary  through  coil  D,  which  we  will 
assume  of  N"  turns.  The  galvanometer  will  again  deflect 
and  we  will  call  this  second  deflection  d'.  Then  if  <£'  is  the 
flux  in  the  specimen  under  test,  we  shall  have 

d  _  4wNN'IS  ,QQ, 

~       '"     ' 


THE  MAGNETIC  FIELD  217 

from  which 

_  d'^NN'IS 
*  '         dN"l 

All  of  the  quantities  on  the  right-hand  side  of  this  equation 
can  be  counted  or  measured  by  convenient  means.  By  the 
ratio  of  two  deflections,  we  thus  find  the  total  flux  in  the 
specimen,  here  shown  as  a  ring,  which  links  the  measuring 
coil  D. 

It  will  be  noted  that  reversing  switches  are  used.  In 
practice,  we  will  throw  each  of  these  switches  in  turn  first 
in  one  direction  and  then  in  the  other,  so  that  the  gal- 
vanometer deflection  will  be  twice  what  it  would  be  for  a 
single  change  of  flux.  The  reason  for  this  is  to  avoid  any 
effect  of  retentivity  in  the  sample,  an  effect  which  will  be 
considered  in  the  next  chapter. 

It  will  be  noted  also  that  B,  D  and  G  are  connected  per- 
manently in  series.  This  is  in  order  to  keep  the  constants 
of  the  circuit  of  the  ballistic  galvanometer  unchanged  while 
it  is  subjected  to  the  two  processes  successively.  The 
deflection  of  the  galvanometer  will  depend  upon  the  re- 
sistance of  the  circuit  to  which  it  is  connected.  For  this 
reason  it  is  necessary  to  keep  the  entire  galvanometer  cir- 
cuit unchanged  while  making  the  calibration  and  the  actual 
measurements  in  order  that  the  deflection  of  the  galvanometer 
may  be  proportional  simply  to  the  two  Values  of  flux  under 
consideration,  one  of  which  we  can  compute  and  the  other 
of  which  we  wish  to  measure.  ' 

Prob.  18-7.  If  in  the  center  of  the  solenoid  of  Prob.  17-7 
there  is  placed  a  coil  B,  as  in  Fig.  109,  having  a  cross-section  of 
1.4  sq.  inches  and  200  turns  of  wire,  what  will  be  the  flux  link- 
ages in  coil  B! 

Prob.  19-7.  If  a  specimen  is  under  test  and  the  galva- 
nometer shows  a  deflection  of  12.3  centimeters  for  the  test 
specimen  and  a  deflection  of  4.2  centimeters  for  the  test  solen- 
oid of  Prob.  18-7,  what  is  the  flux  in  the  specimen? 

Prob.  20-7.  What  is  the  constant  of  the  galvanometer  in 
Prob.  19-7  as  a  flux  meter? 


218      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

75.  The  Toroid.  A  toroid  is  a  solid  in  the  shape  of  a 
doughnut.  For  reasons  explained  later,  this  is  the  shape  of 
the  core  of  the  repeating  coils  used  to  connect  telephone 
circuits  with  one  another.  If  a  core  of  this  form  is  uni- 
formly wound  with  a  magnetizing  winding,  as  shown  in 
Fig.  Ill,  the  resulting  field  may  be  computed  as  follows. 


FIG.  111.     A  toroid,  consisting  of  an  iron  ring.     It  is  magnetized 
by  the  coil. 

By  symmetry,  the  lines  of  force  must  all  be  circles  about 
the  axis  through  the  center  C.  Moreover,  the  magnetizing 
force,  provided  the  toroid  is  uniformly  wound,  must  be  uni- 
form along  any  one  of  these  circles. 

Consider  such  a  circle  of  radius  x,  where  x  lies  between 
the  inside  and  outside  radii  ri  and  r2.  The  line  integral  of 
the  magnetizing  force,  HXJ  around  this  circle  is  equal  to  4?r 
times  the  number  of  abampere-turns  linked.  If  there  are 
N  turns  on  the  toroid  carrying  /  abamperes,  this  gives  the 
equation 

27rxHx  =  4irNI;  (90) 

that  is, 


Since  the  magnetizing  force  around  a  circuit  is  in  this  case 
uniform,  we  may  use  a  core  of  permeability  other  than  unity 


THE  MAGNETIC  FIELD  219 

without  distorting  the  field.  If  the  toroid  is  made  of  iron 
of  permeability  n,  the  flux  density  at  a  radius  x  will  thus  be 
given  by 

&-!.??•  (92) 

X 

The  total  flux,  considering  a  toroid  of  cross-section  A,  will 
then  be  obtained  by  integrating  this  flux  density  over  the 
cross-sectional  area  of  the  core  A.  If  the  thickness  of  the 
core  is  small  compared  with  its  diameter,  we  shall  not  pro- 
duce great  error  if  we  write  the  total  flux  as  equal  to  the 
flux  density  at  a  mean  radius  multiplied  by  the  total  cross- 
sectional  area.  Thus 

2NIA 

4>  =  /*  —  —  •  (93) 

However,  this  equation  is  true  only  in  the  case  where  M  is 
constant  throughout  the  iron  and  for  the  values  used  for 
/.  If  we  wish,  we  may  write  it  in  the  form 

(94) 


Note  that  this  last  formula  is  the  usual  Ohm's  law  for  the 
magnetic  circuit.  If,  as  in  the  case  of  .a  toroid,  we  know 
accurately  the  path  of  the  flux  and  the  cross-section  of  the 
path  at  every  point,  it  is  possible  for  us  to  write  this  equation 
immediately.  The  reluctance  of  the  path  is,  of  course, 
Z/AiA,  where  I  is  the  mean  flux  path,  that  is,  approximately 
the  length  of  the  circumference  at  the  mean  radius  r;  or 

I  =  2<jrr. 

The  total  flux  would  then,  as  we  should  expect,  be  the 
total  magnetomotive  force,  4:r  NI,  divided  by  this  reluctance. 
It  is  important  to  realize,  however,  that  this  is  an  approxi- 
mation, for  the  lengths  of  the  paths  of  the  various  flux  lines 
differ.  We  have  seen  above,  moreover,  that  the  flux  den- 


220      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

sity  varies  inversely  as  the  distance  from  the  axis  of  the 
toroid.  It  is  therefore  not  correct  for  us  to  use  an  average 
value  of  flux  density  and  multiply  by  the  cross-section. 
It  is  permissible  to  do  this  in  problems  where  an  exact 
solution  cannot  be  made  or  where  we  know  that  an  approxi- 
mate solution  is  correct  to  sufficient  engineering  accuracy. 
We  must  be  careful  how  averages  are  used,  however,  and  in 
each  case  be  sure  either  that  an  average  will  give  an  accu- 
rate result  or  that  the  errors  thereby  introduced  are  per- 
missible. It  will  accordingly  be  instructive  for  us  to  com- 
pute the  exact  amount  of  flux  passing  through  a  uniformly 
wound  toroid  under  the  above  conditions  and  compare  it 
with  the  approximate  formula  obtained  by  using  a  mean 
length  of  path  as  above. 


FIG.  112.     A  cross-section  of  a  toroid. 

In  Fig.  112  is  shown  a  cross-sectional  view  of  the  toroid. 
Assume  a  distance  x,  and  an  element  of  this  cross-section 
cut  out  by  the  width  dx.  The  area  of  this  element  is  evidently 


2  Vp2  -  (r  -  xY  dx.  (95) 

Since  we  have  found  that  the  flux  density  at  this  distance 
from  the  center  line  x  is 

ft  =  ^,  (96) 


THE  MAGNETIC  FIELD  221 

the  total  flux  passing  through  this  element  of  the  cross- 
section  is 


NI 


d<}>  =  4M  —  Vp2  -  (r  -  x}2  dx.  (97) 

x 

We  may  thus  obtain  the  total  flux   by  integrating  this 
expression  from  x  =  r\  to  x  =  r^\  that  is, 


(98) 

The  evaluation  of  this  integral  is  somewhat  involved,  but 
it  may  be  worked  out  when  ^  is  a  constant  by  means  of  a 
table  of  integrals.*  Evaluating,  inserting  the  limits  and 
making  a  few  simple  reductions,  it  becomes 

W  (r  -  Vr"^=~PT).  (99) 

Inserting  this  value  in  our  expression  for  the  total  flux,  we 
obtain  _ 

<j>  =  ^NlTr  (r  -  Vr*  -  P2).  (100) 

This  is  the  exact  expression  for  the  total  flux  in  a  toroid, 
but  here  again  ^  must  be  a  constant. 

When  p  is  small  compared  with  r,  that  is,  when  the  toroid 
is  of  large  diameter  and  correspondingly  small  cross-section, 
we  may  obtain  an  approximate  formula,  by  expanding  the 
radical  by  the  binomial  theorem,  thus: 

<\ 

(r2  _  p2)i/2  =  r  _  1/2^-+  ----  (101) 

If  we  neglect  terms  beyond  the  second  in  this,  our  formula 
becomes 

(102) 


which,  using  the  abbreviations 

1  =  2irr 
*  See  Pierce,Table  of  Integrals,  187,  183,  160 


222      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

and  A  =  wp2, 

becomes 

_4*NI 

0=     ~1T  (103) 

M^ 

the  same  formula  which  we  derived  for  the  toroid  from 
the  simple  assumption  that  the  reluctance  could  be  com- 
puted from  the  length  of  path  at  mean  radius. 

In  cases  where  p  is  not  negligible  in  size  compared  with 
r,  we  must  use  the  exact  formula 


r  -  W2  -  p2).  (104) 

A  good  idea  may  be  formed  of  the  approximation  in- 
volved in  using  the  approximate  formula  by  working  out 
an  example. 

Assume  a  toroid  of  mean  radius  4  centimeters,  the  radius  of 
the  cross-section  being  2  centimeters.  We  shall  then  have 

r  =4, 

'  :  L  <105> 

A     =    47T. 

Using  the  approximate  formula  for  the  flux  in  this  case,  it 
becomes 

(106) 


The  exact  formula,  on  the  other  hand,  gives 

<*»  =  47TAT/M  (4  -  \/42  -  22).  (107) 

This  becomes,  upon  reduction 

(1.07).  (108) 


We  thus  see  that  an  error  of  about  7%  would  be  introduced 
by  assuming  that  the  approximate  formula  was  nearly 
enough  correct.  This  approximate  formula  is  based  upon 
an  average.  Thus  averages  can  be  used  only  where  linear 
relations  hold,  that  is,  where  averages  will  yield  exact 


THE  MAGNETIC  FIELD  223 

values;  or,  on  the  other  hand,  where  the  amount  of  error 
introduced  by  including  an  average  can  be  estimated  and 
tolerated. 

In  this  latter  case,  if  the  core  of  the  toroid  were  of  iron, 
say  of  permeability  1000,  an  error  of  even  6%  in  the  formula 
would  be  tolerable,  for  the  reason  that  the  value  of  n  would 
in  many  cases  be  in  doubt  to  this  extent  and  the  formula 
holds  only  when  /*  is  constant.  If,  however,  such  a  toroid 
were  being  used  for  finding  an  exact  value  for  the  permea- 
bility of  a  sample  of  iron,  this  6%  error  might  be  very 
serious.  It  is  thus  evident  that  the  question  of  whether 
or  not  an  error  can  be  tolerated  depends  entirely  upon  the 
conditions  under  which  the  computation  is  being  made. 

Referring  back  to  Fig.  Ill,  suppose  we  assume  the  value 
of  x  less  than  r\.  The  line  integral  of  H  around  the  cir- 
cumference of  this  radius  must  then  be  zero,  for  such  a 
circumference  does  not  link  any  circuit  carrying  current. 
By  symmetry,  therefore,  H  all  the  way  along  the  circum- 
ference must  be  zero.  This  means  practically  that  outside 
of  a  uniformly  wound  toroid  there  is  no  field  whatever. 
In  other  words,  such  an  arrangement  has  no  leakage  flux. 
For  this  reason  and  others,  cores  of  this  sort  are  used  in  the 
repeater  coils  of  telephony,  where  a  leakage  flux  would 
cause  cross-talk,  which  is  naturally  to  be  avoided  where- 
ever  possible.  In  a  practically  wound  toroid,  there  will, 
however,  always  be  a  certain  minute  amount  of  leakage  flux, 
since  it  is  impossible  to  wind  a  coil  uniformly.  The  wires 
must  have  insulation,  which  means  that  as  we  pass  around 
the  circumference  of  the  core,  we  encounter  first  a  space 
carrying  current  and  then  a  space  of  insulation.  In  order 
to  have  an  absolutely  uniformly  wound  coil,  the  insulation 
would  have  to  be  entirely  eliminated  and  the  wires  made 
rectangular  in  form  and  closely  spaced,  as  for  instance  is 
shown  in  the  cross-sectional  view  of  Fig.  113.  Such  an 
arrangement  is  known  as  a  "  current  sheet."  In  a  practical 
arrangement  with  round  wires,  as  shown  in  Fig.  114,  there 


224      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

will  always  be  a  certain  amount  of  leakage  flux  around  each 
individual  wire. 

Prob.  21-7.  A  cast-iron  toroid  with  the  dimensions  r  =  3 
inches,  p  =  0.25  inch,  has  2000  turns  of  wire  wound  uniformly 
about  the  core.  If  the  current  is  0.1  ampere,  what  will  be  the 
flux  in  the  core?  See  Fig.  111. 


^ 

FIG.  113.     Closely  wound  coils  are  considered  to  be  constructed  of 
rectangular  wire  with  thin  insulation  between  turns. 

Prob.  22-7.  How  many  turns  would  be  required  to  produce 
a  flux  density  of  65.000  lines  to  the  square  inch  in  the  cast- 
iron  ring  of  Prob.  21-7? 


FIG.  114.     Coils  are  usually  constructed  of  round  wire  and  some  leak- 
age takes  place  between  turns. 

Prob.  23-7.  (a)  What  is  the  magnetomotive  force  acting 
upon  each  line  of  force  in  the  ring  of  Prob.  22-7? 

(6)  What  is  the  magnetizing  force  in  ampere-turns  per  inch 
acting  upon  the  lines  of  force  at  the  outer  edge? 

(c)  At  the  inner  edge? 

(d)  At  the  center? 

(e}  Compute  the  average  magnetizing  force  and  compare 
it  with  the  magnetizing  force  at  the  center. 

76.  Flux  About  a  Conductor  in  a  Slot.  We  have  now 
examined  magnetic  circuits  in  two  forms;  first,  those  in 
which  the  magnetic  circuit  was  entirely  in  iron;  second, 
those  in  which  the  magnetic  circuit  was  entirely  in  air  or 
other  non-magnetic  materials.  We  have  even  examined 
a  circuit  in  which  a  short  air  gap  was  introduced  into  a 


THE  MAGNETIC  FIELD 


225 


circuit  otherwise  of  iron.     In  all  of  these  cases  it  has  been 
possible  to  arrive  at  accurate  solutions. 

There  is  another  class  of  problems,  however,  which  in 
general  are  more  difficult  and  require  approximations  for 
their  solutions.  These  involve  cases  where  the  magnetic 
circuit  is  partly  in  iron  but  to  a  considerable  extent  in  air. 


iff* 


FIG.  114a.  Section  of  a  motor  frame  showing  method  of  installing  coils 
in  slots.     The  General  Electric  Co. 

In  such  cases  the  reluctance  of  the  magnetic  circuit  is  usually 
almost  entirely  in  the  air  path.  This  is  because  of  the  fact 
that  there  is  such  a  great  difference  between  the  permea- 
bility of  air  and  the  permeability  of  iron.  If  the  length  of 
the  air  path  is  a  considerable  percentage  of  the  length  of 
the  total  path  of  the  flux,  the  reluctance  of  the  path  of  iron 
may  usually  be  entirely  neglected.  In  cases  of  this  sort,  the 
difficulty  of  computation  arises  by  reason  of  the  fact  that 
it  is  usually  difficult  to  map  out  the  path  of  the  flux  lines 
and  accordingly  to  compute  the  reluctance  of  the  air 
path. 


226      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


<t-c-> 

T 

\ 

L_ 

w 

%%& 

FIG.  115.  A  conductor 
lying  at  the  bottom  of 
a  slot  in  iron. 


One  important  problem  of  this  sort  is  that  of  a  conductor 
carrying  current  and  imbedded  in  a  slot  in  iron.  Such  an  ar- 
rangement is  shown  in  Fig.  114a,  each  slot  of  the  armature 
core  containing  two  conductors.  Fig.  115  is  a  drawing  of  one 
conductor  in  a  slot,  the  cross-hatched  area  being  the  con- 
ductor of  rectangular  form,  which 
we  will  assume  carrying  a  current 
of  /  abampercs.  The  outlined  area 
is  of  iron.  This  figure  is  a  cross- 
section,  the  conductor  lying  in  a 
trough  in  the  iron.  Such  an  ar- 
rangement as  this  is  found  in  al- 
most all  electrical  machinery.  The 
amount  of  flux  produced  by  the 
conductor  is  of  importance  in  con- 
sidering the  regulation  of  alternators,  in  determining  the 
proper  commutation  of  direct-current  machines  and  so  on. 
We  will  hence  examine  the  methods  of  computing  the 
amount  of  flux  produced  by  such  a 
conductor. 

The  flux  will  surround  the  con- 
ductor somewhat  as  shown  in  Fig. 
116.  This  flux  may  be  divided  into 
two  parts,  a  part  A  which  passes 
directly  across  from  tooth  to  tooth, 
and  a  part  B  which  passes  around 
between  the  ends  of  the  teeth.  We 
will  first  examine  the  part  A. 

For  a  flux  line  which  crosses  the 
slot  above  the  conductor,  as  shown 
m,  Fig.  116,  the  computation  is 
simple.  Since  the  magnetizing  force 

is  almost  entirely  used  up  in  the  reluctance  of  the  air  path, 
we  may  write  the  line  integral  of  H  around  this  path  as 
simply  the  product  of  H  times  the  width  of  the  gap  b. 
This  line  integral  is  equal  to  the  magnetomotive  force  of 


FIG.  116.  The  flux  distri- 
bution about  a  con- 
ductor in  a  slot  in  iron. 


THE  MAGNETIC  FIELD 


227 


the  conductor,  which  will  be  4?r  7,  since  there  is  only  one 
turn  involved.  We  have,  therefore, 

Hb  =  4T  /  gilberts.  (109) 

In  the  air,  of  permeability  1,  the  flux  density  will  be  equal 
to  H,  and  hence  we  have  for  the  flux  density  in  the  portion 
of  the  space  above  the  top  of  the  conductor  and  below  the 
top  of  the  slot  a  flux  density 


D      4x7 

B  =  —7-—  gausses. 


(110) 


Consider  now  a  flux  path  such  as  2  in  Fig.  117.  This 
crosses  the  conductor  at  a  distance  x  above  the  bottom  of 
the  conductor.  The  line  integral  of  this  path  is  the  same  as 


<  —  b—* 

1 

-> 

*-*? 

/"" 

1 

1 

1 

"• 

ri 

14— 

/ 

k>- 

!  / 

FIG. 


117.     Distribution  of  flux  density  about  a  conductor  lying  in 
a  slot.     Tooth-tip  flux  is  not  included  in  this  figure. 


before,  namely  Hb.  This  path,  however,  does  not  link  the 
entire  current,  but  only  that  part  of  the  gurrent  in  the  con- 
ductor lying  below  the  distance  x,  that  is,  the  fraction  x/h 
of  the  total  current.  We  have,  therefore,  for  this  path  the 
formula 

H b  =  4w  r  7  gilberts, 
which  gives  immediately 


(112) 


r>  T 

B  =  -T-  r  I  gausses. 
o  h 


The  flux  which  passes  across  the  actual  conductor  thus  has 
a  density  which  is  proportional  to  the  distance  x,  that  is,  to 


228      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


the  distance  above  the  bottom  of  the  conductor.  If  we 
plot  the  flux  density  against  the  distance  above  the  bottom 
of  the  slot,  we  shall  accordingly  obtain  a  figure  such  as  fs 
shown  in  the  right-hand  half  of  Fig.  117. 

The  area  between  this  curve  and  the  vertical  axis  is 
evidently  the  total  amount  of  flux  per  centimeter  length  of 
slot  passing  directly  across  from  tooth  to  tooth.  We  there- 
fore obtain  for  the  portion  A  of  the  total  flux 

.,4*7         4rJ  h 
*A      ~T  b    2 

=  —r—  1 1  +  -  j  lines  per  cm.  length  of  slot. 

=  — T—  (  t  +  -  J  linkages  per  cm.  length  of  slot.  (113) 

In  order  to  compute  the  amount  of  flux,  E,  passing 
between  the  ends  of  the  teeth,  it 
will  be  necessary  to  assume  some 
shape  for  the  path  of  these  lines 
which  seems  reasonable  and  which 
admits  of  computation.  A  path  such 
as  shown  in  Fig.  118  is  a  reasonable 
assumption.  The  flux  E  at  a  distance 
from  the  edge  of  the  tooth  y  is  as- 
sumed to  pass  in  a  path  made  up  of 
two  arcs  of  circles  and  a  straight  line. 
The  length  of  such  a  path  is 


f'~ 

«  —  6— 

>- 

~v^ 

h* 

i 
i 

j 

i 

B 

i 

FIG.    11 

8.      T 

h 

e    distn 

bution  of  the  tooth-tip 


o  +  Try 
and  since  this  path  links  the  entire 


flux  density  set  up  by  flux>  we  may  write 
a  conductor  in  a  slot.  jjr^    \    ^ 

which  gives 


B  = 


b  +  Try 


gausses. 


(U4) 


(115) 


This  distribution  of  flux  density  is  shown  by  the  diagram 


THE  MAGNETIC  FIELD  229 

of  Fig.  118.    The  total  flux  passing  between  the  teeth  will 
be  the  area  under  this  curve,  or 

r47r  i 
j—7-  —  dy  maxwells,  (116) 

0  +  Try 

which  integrated  gives 

fa  =  47  log  (6  +  T2/)?.  (117) 

Jo 

This,  when  the  limits  have  been  substituted,  becomes 

fo  =  47  log  Mil?  maxwells.  (118) 

The  total  flux  produced  by  the  conductor  A  will  then  be 
the  sum  of  <J>A  and  4>s9  or 


4irl  /.   .  h\       A  T  ,     b  +  ira  ,. 
=  -T-  (  t  +  2  )  +  4^  loS  —  £  —  lmes  Per  centi- 

meter length  of  slot.  (119) 

We  have  thus  derived  a  formula  for  most  of  the  flux  pro- 
duced by  a  conductor  in  a  slot.  The  flux  thus  computed  is 
confined  to  the  air  and  to  the  iron  of  the  core  and  the 
adjacent  teeth.  Additional  flux  may  be  set  up  in  the  core 
and  in  teeth  somewhat  removed  from  the  slot  considered. 
Such  problems  are  of  common  occurrence  in  electrical  design 
and  calculation.  It  will  generally  be  fotmd  that  they  can 
be  solved  in  the  manner  of  the  above  problem  to  obtain  at 
least  an  approximate  result,  by  assuming  a  reasonable  path 
for  the  flux  as  a  basis  for  the  calculation. 

Prob.  24-7.     Taking  the  following  dimensions  as  applying 
to  Fig.  115: 

a  =  0.75  inch 
b  =  0.80  " 
c  =  0.75  " 
d  =  2.50  " 
h  =  1.15  " 
k  =  0.05  " 
s  =  1.20  " 
t  =  1.30  "  , 


230     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

plot  the  flux  density  curve  and  determine  the  total  flux  (ex- 
cluding flux  passing  between  the  tips  of  the  teeth)  for  a  current 
in  the  conductor  of  500  amperes. 

Prob.  25-7.  With  the  data  of  Prob.  24-7,  plot  the  flux  pass- 
ing between  the  ends  of  the  teeth  and  find  the  total  value. 

Prob.  26-7.  Compute  the  total  value  of  the  flux  set  up  by  a 
current  of  500  amperes  in  a  conductor  in  the  top  of  a  slot. 
Slot  and  conductor  have  the  dimensions  of  Fig.  115,  k  being 
the  distance  which  the  top  of  the  conductor  is  below  the  ends 
of  the  teeth,  and  the  values  of  s  and  t  being  interchanged. 

77.  Introduction  of  Iron  into  a  Uniform  Field.  Engineers 
occasionally  have  to  deal  with  a  problem  in  which  a 
piece  of  iron  is  introduced  into  a  mag- 
netic field  initially  approximately  uni- 
form. Such  a  problem  is  presented,  for 
example,  by  an  iron-core  solenoid  as 
shown  in  Fig.  119.  This  is  a  cross- 
sectional  diagram,  and  shows  simply  a 
coil  of  wire  into  the  center  of  which  we 
FIG.  119.  Plunger  have  Pr°Jected  an  iron  plunger.  In  such 
type  of  solenoid.  sectional  drawings  it  is  usual  to  represent 
the  cross-sectional  area  of  the  winding 
by  a  double  cross-hatch.  When  current  is  applied  to  this 
winding,  the  iron  plunger  will  be  drawn  toward  the  center 
of  the  coil.  Solenoids  are  often  used  in  electrical  appliances 
in  the  place  of  electromagnets  in  which  the  core  is  station- 
ary and  an  armature  moves,  because  of  the  longer  distance 
of  movement  which  can  be  obtained  with  the  solenoid 
arrangement.  Relays  such  as  are  used  in  a  power  station 
for  opening  the  main  oil  switches,  in  case  of  short  circuit, 
are  usually  solenoids.  The  over-load  release  of  a  direct- 
current  starting  box  is  another  example. 

These  appliances  present  problems  which  are  usually 
difficult  of  solution  on  account  of  the  doubt  existing  as  to 
the  path  of  the  magnetic  flux.  We  have  seen  that  an  air- 
core  solenoid  may  be  fairly  accurately  computed,  on  the 


THE  MAGNETIC  FIELD  231 

simple  assumption  that  all  of  the  reluctance  of  the  path  of 
the  flux  lies  within  the  winding  and  that  the  reluctance  of 
the  return  path  may  be  neglected.  When,  however,  iron  of 
high  permeability  fills  part  of  this  internal  path,  the  outside 
reluctance  can  no  longer  be  neglected  in  comparison  with 
the  reluctance  inside  of  the  coil.  In  order  to  solve  such  an 
arrangement,  it  is  therefore  necessary  to  map  out  the  actual 
flux  paths. 

In  engineering  calculations  it  is  usually  sufficiently  ac- 
curate to  map  out  these  paths  in  the  same  manner  as  was 
done  in  the  last  section  in  the  case  of  a  conductor  in  a  slot. 
The  accurate  solution  of  a  problem  such  as  an  iron-core 
solenoid  involves  difficult  mathematics  and  is  seldom  under- 
taken by  engineers.  In  fact,  the  more  accurate  solution 
must  always  be  based  upon  some  assumption  in  regard  to 
the  permeability  of  the  material  used.  For  this  reason,  the 
mathematically  accurate  solution,  based  as  it  is  upon  an 
incomplete  premise,  is  likely  to  be  nearly  as  much  in  error 
as  a  solution  practically  based  upon  reasonable  assumption 
in  regard  to  simple  paths  for  the  flux  lines. 

It  will  not  be  worth  while  for  us  to  analyze  such  problems 
completely,  for  to  treat  such  cases  with  any  great  degree  of 
thoroughness  would  require  a  complete  book.*  There  are, 
however,  several  simple  propositions  in  regard  to  the  paths 
taken  by  flux  lines  which  if  kept  in  mind  will  aid  the  engineer 
in  making  sufficiently  accurate  estimates.  Such  simple 
propositions  taken  together  with  engineering  experience 
will  usually  enable  him  to  solve  his  problems  with  sufficient 
engineering  accuracy.  The  usual  practical  method  of 
designing  a  solenoid,  in  fact,  is  to  build  several  of  approxi- 
mately the  size  and  shape  desired,  measure  their  charac- 
teristics and  make  a  final  design  based  upon  this  experience. 
Such  a  proceeding  is  likely  to  be  quicker  and  less  expensive 
than  an  attempted  complete  mathematical  analysis.  An 
accurate  knowledge  of  some  of  the  simple  rules  in  regard 
*  For  example,  "Solenoids,"  by  Underbill, 


232       PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

to  the  flux  paths  is,  however,  of  great  service  in  analyzing 
and  comparing  the  results  of  such  cut-and-try  tests. 

In  the  first  place,  it  must  be  realized  that  we  deal  always 
with  magnetic  circuits,  just  as  we  deal  always  with  electric 
circuits.  We  write 


B  =  pi!  gausses 

as  a  general  law,  and  we  have  found  that  in  the  neighborhood 
of  a  long  conductor  carrying  current  we  may  also  write 

27 

H  =  —  gilberts  per  centimeter,  (120) 

which  tells  us  that  at  a  distance  of  1  centimeter  from  such  a 
center  of  a  long  wire  carrying  a  current  of  1  abampere,  the  mag- 
netizing force  will  be  2  gilberts  per  centimeter.  This  does 
not  mean,  however,  that  in  the  case  of  a  wire  situated  in  an 
iron  trough  of  semicircular  cross-section,  as  shown  in  Fig. 
120,  we  can  find  the  flux  density  at  a  point  in  the  iron  trough 
by  multiplying  this  value  of  H  by  the  permeability  of  iron. 

1  Abampere 


0 

3 

(b) 

120.     A  copper  conductor  in  an  iron  trough. 

The  permeability  of  iron  may  be  1000,  but  the  flux  density  in 
the  iron  will  not  by  any  means  be  2  times  1000.  Such 
a  fallacy  is  one  often  committed  by  engineers  who  lack 
experience.  It  is  exactly  the  same  thing  as  if  we  dealt  with 
an  electric  circuit  consisting  of  a  battery  giving  an  electro- 
motive force  of  1  volt,  with  an  internal  resistance  of  0.1 
ohm,  applied  to  an  external  resistance  of  10  ohms.  The 
electromotive  force  in  the  battery  itself  is  1  volt,  and  its 
internal  resistance  0.1  ohm.  We  might  say  that  the  con- 
ductivity of  the  battery  is  10  mhos.  It  would  be  foolish, 


THE  MAGNETIC  FIELD  233 

however,  to  assume  that  the  current  in  the  battery  is  10 
times  1  ampere.  We  know  that  we  must  treat  the  entire 
electric  circuit.  In  the  same  way,  we  must  treat  an  entire 
magnetic  circuit. 

If  the  trough  in  Fig.  120  is  removed,  we  know  that  the  flux 
density  at  a  distance  of  1  centimeter  from  the  center  of  the 
wire  will  be  2  gausses.  The  total  magnetizing  force  is  4ir  gil- 
berts, and  the  length  of  the  magnetic  circuit  2ir  centimeters. 
This  magnetizing  force  is  uniformly  applied  along  the  mag- 
netic circuit,  and  hence  we  have  everywhere  the  magnet- 
izing force  of  2  gilberts  per  centimeter.  Since  the  per- 
meability of  air  is  unity,  the  resulting  flux  density  will  then 
be  2  gausses.  Moreover,  if  we  have  a 
complete  cylindrical  sheath  about  the 
wire,  composed  of  iron,  as  indicated  in 
Fig.  121,  the  permeability  of  iron  being 
taken  as  1000,  we  may  say  immediately 
that  the  flux  density  in  this  sheath  will 

be  2000  gausses.     We  deal  again  with  a 

,.  ,.         FIG.  121.     A  copper 

complete  magnetic  circuit,  the  magnetiz-        conductor  in  an 

ing  force  is  uniformly  distributed  along        jron  sneath. 
this  circuit,  and  therefore  the  flux  density 
can  be  found  by  multiplying  this  uniform  magnetizing  force 
by  the  permeability. 

In  the  arrangement  shown  in  Fig.  120,  however,  a  com- 
plete magnetic  circuit  is  made  up  partly  of  iron  and  partly  of 
air.  The  iron  part  is  of  low  reluctance  and  the  air  part  of 
high  reluctance.  The  total  magnetomotive  force  will  hence 
be  used  up  almost  entirely  in  the  air  part  of  the  circuit. 
The  magnetomotive  force  acting  across  the  iron  will  be  very 
small.  The  magnetizing  force  in  the  iron,  that  is,  H,  will 
thus  also  be  small.  If  we  can  find  this  small  magnetizing 
force,  and  multiply  it  by  the  permeability,  we  shall  have 
the  resulting  flux  density  in  the  iron.  The  computation  of 
the  actual  value  of  H  is,  however,  very  difficult. 

Another  illustration  of  this  principle  is  of  importance. 


234      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

When  there  is  no  iron  about,  we  have  a  very  uniform  field, 
due  to  the  earth.  The  intensity  of  this  field  is  about  0.6 
gauss.  We  have  everywhere  acting  a  magnetizing  force  of 
0.6  gilbert  per  centimeter.  Suppose  we  place  a  cannon  ball 
of  permeability  3000  in  such  a  field.  If  to  find  the  flux  den- 
sity we  simrjly  multiplied  M  by  H,  we  should  obtain  1800 

gausses  for  the  flux  density  in  the  cannon 

ball.     With  such  a  flux  density,  the  ball 
;     would  be  very  highly  magnetized  indeed. 

In  fact,   if  two  of  them  were  placed  side 

* by  side  on  the  floor,  each  thus  magnetized, 

FIG.  122.  A  uni-  they  would  roll  around  violently.  We  know, 
form  parallel  of  course,  that  such  a  thing  does  not  occur. 
field-  The  resulting  magnetization  in  the  iron  ball 

can  with  difficulty  be  detected.  The  ex- 
planation is  simple.  The  iron  ball  is  introduced  into  a 
uniform  field,  and  since  its  reluctance  is  very  low,  there  is 
very  little  reluctance  drop  from  one  end  of  it  to  the  other. 
The  resulting  H  in  its  interior  is  ac- 
cordingly very  small.  The  field  before 
the  introduction  of  the  ball  is  shown  in 

Fig.  122.     After  the   introduction  of  the 

ball,  it  is  much  as  shown  in  Fig.  123.     A " 

few  of  the  flux  lines  are  diverted  and  pass     FlQ  123 
through    the    ball.      The  flux  density  at        fieid  Of  Fig.  122 
its  interior  will  be  about  0.4  gauss.     This        after    an    iron 
flux  density  is  obtained  by  multiplying  /z         ball    has    been 
by  the   resulting  magnetizing  force,  and        introduced. 
not  by  the  original  force. 

An  analogy  will  show  the  reason  why  the  flux  density  in 
the  cannon  ball  merely  doubles.  Suppose  that  we  take  a 
square  yard  of  thin  rubber  sheet,  attach  sticks  to  the  two 
opposite  sides  and  stretch  it.  We  have  a  uniform  strain 
in  the  rubber  which  can  be  represented,  if  we  wish,  by  a 
series  of  parallel  strain  lines.  Now  let  us  cut  a  circular  hole 
in  the  rubber.  In  this  hole  the  resistance  of  the  air  to  stress 


THE  MAGNETIC  FIELD  235 

is  zero,  and  yet  it  is  not  deformed  very  much.  The  hole 
becomes  elongated  so  that  its  length  is  about  twice  its 
width.  It  stretches,  in  fact,  until  there  is  no  stress  left  in 
its  interior.  The  resistance  of  the  sheet  to  strain  is  analo- 
gous to  the  reciprocal  of  the  permeability.  The  hole  cor- 
responds to  a  cannon  ball  of  infinite  permeability.  It  will 
be  seen  from  this  illustration  why,  if  the  permeability  of 
the  ball  is  many  times  that  of  air,  almost  the  same  effect 
will  be  produced  as  if  the  permeability  were  actually  infinite. 

During  the  recent  war,  one  method  that  was  tried  for 
detecting  the  presence  of  enemy  submarine  boats  was  to 
observe  by  various  means  the  distortion  in  the  earth's 
magnetic  field  produced  by  the  shells  of  the  submarine. 
The  permeability  of  the  material  of  the  submarine  boats' 
plates  was  unknown,  and  also  their  thickness.  Practically 
the  same  effect  on  the  earth's  field  was  obtained,  however, 
as  if  the  submarine  had  been  solid  and  of  infinite  permea- 
bility. The  effect  was  small,  however,  and  could  be  de- 
tected at  distances  of  a  few  hundred  yards  only  with  ex- 
treme difficulty. 

It  will  be  noted  in  Fig.  123  that  the  flux  lines  where  they 
enter  and  leave  the  ball  are  drawn  perpendicular  to  the 
surface.  It  is  a  general  proposition  that  flux  lines  entering 
or  leaving  iron  of  high  permeability  will  pass  in  or  out  almost 
exactly  perpendicular  to  the  surface,  Tliis  fact  is  of  great 
assistance  in  mapping  out  magnetic  fields  where  iron  is 
involved.* 

The  reason  that  this  must  be  true  can  readily  be  made 
clear.  In  Fig.  124,  suppose  that  the  area  above  the  dividing 
line  is  air,  and  that  below  the  line  is  iron  of  high  permeability. 
H  inside  the  iron  is  very  small.  It  is  equal  to  5/M,  and 
since  ju  is  very  large,  the  resulting  H  can  practically  be 

*  When  the  flux  density  in  the  iron  is  very  high,  and  in  the  air  very 
small,  this  rule  has  exceptions;  for  example,  on  armature  teeth  under 
certain  conditions.  Such  cases  are,  however,  extreme.  See  Rogofsky, 
Archiv.  Fur.  Elek.,  Band  9,  1920. 


236      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

neglected  in  comparison  with  the  value  of  H&,  for  the  air. 
Consider  a  closed  path  drawn  in  the  magnetic  field  as  shown 
by  the  dotted  line.  The  line  integral  of  H  around  this 
path  will  evidently  be  zero.  The  line  integral  across  ab  is  zero, 
since  the  path  here  is  perpendicular  to  the  flux.  The  part  of 


FIG.   124.     The  correct  method  of  representing  the  angle  at  which 
magnetic  lines  enter  iron  from  air. 

the  line  integral  along  cd  is  practically  zero,  for  this  path  is 
entirely  in  iron.  H  in  the  iron,  we  have  seen,  is  so  small 
it  can  be  neglected  without  serious  error.  The  integral 
along  the  paths  be  and  da  will  cancel,  for  these  paths  are 


FIG.  125.     The  lines  are  incorrectly  shown  entering  the  iron  at  less  than 
right  angles  to  the  surface. 

equal  in  length  and  opposite  in  direction.  The  line  integral 
around  this  closed  path  of  H  is  accordingly  zero;  and  this 
is  as  it  should  be,  for  the  path  links  no  conductors  carrying 
current. 

Suppose,  however,  that  we  consider  Fig.  125,  where  the 
flux  lines  are  incorrectly  drawn  entering  the  surface  at  an 
angle  much  less  than  90°.  It  will  always  be  possible  to 
draw  a  path  such  as  efg,  around  which  the  line  integral  of 
H  is  not  zero.  The  integral  along  ef  and  fg  is  zero  as  before. 
That  along  the  path  ge,  however,  is  not  zero.  The  total 


THE  MAGNETIC  FIELD 


237 


line  integral  is  therefore  not  zero.  We  know  this  to  be  in- 
correct, for  the  line  integral  around  such  a  closed  path  must 
be  zero  if  it  links  no  conductors  carrying  current.  This  ar- 
rangement, therefore,  must  be  incorrect,  and  it  is  not  gener- 
ally possible  to  have  flux  lines  entering  a  material  of  high  per- 
meability from  a  material  of  low  permeability  at  an  acute 
angle  with  the  surface.  In  drawing  flux  diagrams  in  which 
air  and  iron  are  both  involved,  we  should  be  careful  to  draw 
the  flux  lines  always  perpendicular  where  they  enter  the  iron 
surface  from  the  air,  and  where  they  leave  the  iron  for  the  air. 
There  is  one  further  principle  which  will  assist  in  the 
proper  mapping  of  flux  fields  and  in  making  approximate 
estimates  of  the  reluctance  of  air  gaps,  etc. 


FIG.  126. 


The  lines  leave  and  enter  the  iron  perpendicularly 
to  the  surface. 


Fig.  126  shows  a  pair  of  iron  pieces  which  compose  a  part 
of  a  magnetic  circuit,  with  an  air  gap  between  them.  The 
field  in  this  air  gap  is  drawn,  fringing  widely  at  the  edge  of 
the  gap.  It  is,  of  course,  necessary  to  draw  only  a  few  lines. 
The  flux  density  in  the  air  gap  may  amount  to  thousands  of 
lines  per  square  centimeter,  but  only  a  few  of  these  need  be 
drawn  to  show  the  direction  of  the  field.  If  these  are  drawn 
at  regular  intervals,  they  divide  the  gap  between  the  iron  pieces 


238      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

into  a  number  of  unit  volumes.  Since  these  filaments  are 
all  in  parallel,  there  must  be  the  same  reluctance  drop  across 
all  of  them.  Neglecting  the  small  reluctance  drop  in  the 
iron  pieces  themselves,  the  reluctances  of  these  air  filaments 
must  all  be  equal.  They  should  therefore  be  drawn  in  such  a 
manner  that  the  quotient  of  the  length  divided  by  the  cross- 
sectional  area  of  each  filament  is  the  same.  If  we  are  con- 
sidering fringing  in  one  direction  only,  that  is,  if  in  Fig. 
126  the  iron  pieces  extend  a  long  distance  in  the  direction 
perpendicular  to  the  paper,  this  means  that  the  width  of 
each  space  between  lines  would  on  the  average  be  propor- 
tional to  the  length  of  the  lines.  If  filament  b  is  twice  as 
long  as  a,  its  width  would  thus  be  twice  as  great  as  the 
width  of  a.  Filament  c,  which  is  about  three  times  as  long 
as  a,  would  also  be  about  three  tunes  as  wide. 

After  we  have  mapped  out  the  field  in  this  manner,  the 
total  reluctance  of  the  air  gap  may  be  found  by  computing 
the  reluctance  of  one  filament,  and  dividing  by  the  number 
of  filaments.  Thus  in  Fig.  126,  suppose  filament  a  is  2 
centimeters  long  and  0.3  centimeter  wide.  Suppose  the 
iron  pieces  are  10  centimeters  broad  in  the  direction  per- 
pendicular to  the  paper.  The  reluctance  of  filament  a  will 

=  0.667  oersted. 


03x 

There  are  17  such  filaments  in  the  figure,  if  we  do  not  con- 
sider the  fringing  around  the  ends  of  the  pieces.  The  total 
reluctance  of  the  air  gap  is  accordingly 


R  =  =  o.039  oersted. 

Compare  this  with  the  reluctance  of  the  air  gap  on  the 
assumption  that  the  flux  goes  straight  across  without  any 
fringing. 

R'  =  o  ^  m  =  °-067  oersted. 
o  X  1U 


THE  MAGNETIC  FIELD 


239 


The  actual  reluctance  thus  has  the  ratio  to  this  uncorrected 
reluctance  of 

R  =  0.039 

R'      0.067 


59%. 


The  fringing  in  this  case  decreases  the  reluctance  of  the  air 
gap  to  a  little  over  half  of  what  it  would  be  if  there  were  no 
fringing. 

As  a  rough  rule  to  cover  this  case,  we  have  previously 
proposed  to  add  to  the  width  of  the  flux  path  an  amount 
equal  to  its  length.  Applying  this  rule  to  our  present  gap, 
we  should  have  a  length  of  2  centimeters  and  a  width  of 
3  +  2  or  5  centimeters.  The  reluctance,  disregarding  the 
end  fringing,  would  then  be 


5-xTo  -  a°40> 


R"  = 


which  is  quite  close  to  the  value  which  we  obtained  by  map- 
ping the  field.     This  rule  is  usually  sufficiently  accurate 


FIG.  127. 


The  lines  enter  and  leave  the  poles  at  right-angles 
to  the  surface. 


for  short  air  gaps.  In  cases  where  the  air  gap  is  not  simple, 
or  where  it  is  long,  the  above  process  of  mapping  the  field 
will  be  found  to  be  of  service. 

Sometimes  we  must  consider  the  fringing  in  two  directions. 
Consider    the    field    shown   in   Fig.    127,    where   the   pole 


240      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

pieces  are  circular  in  cross-section.  In  this  case  we  must 
remember  that  the  filaments  of  air  bounded  by  the  flux  lines 
we  draw  have  two  dimensions,  and  the  second  dimension 
varies  for  different  filaments.  The  lines  we  have  drawn 
show  a  cross-section  of  the  field.  We  may,  if  we  like,  con- 
sider the  air  gap  to  be  divided  into  surfaces  of  revolution 
such  as  would  be  obtained  by  revolving  the  figure  shown 
about  its  axis.  The  total  cross-sectional  area  of  a  filament 
thus  generated  would  be  proportional  to  the  length  of  its 
boundary.  The  resulting  figure  will,  of  course,  not  show  a 
proper  cross-sectional  view  of  the  distribution  of  flux  lines, 
but  it  may  be  more  convenient  as  a  means  of  computation. 
78.  Magnetic  Poles  and  Pole  Strength.  The  subjects 
of  magnetic  poles  and  magnetic  pole  strength  have  been 
purposely  left  out  of  the  above  discussion.  In  many  texts 
it  will  be  found  that  some  of  the  above  matters  are  computed 
in  a  different  manner,  by  finding  the  strength  of  the  poles  and 
assuming  that  they  contribute  to  a  resulting  magnetizing 
force.  This  method  has  long  been  in  use  by  physicists. 
It  is  historically  much  older  than  the  method  that  is  here 
presented.  However,  engineers  now  depend  almost  en- 
tirely upon  the  laws  and  methods  of  computation  which 
have  been  used  in  this  chapter.  Magnetic  circuits  are 
treated  by  them  in  exactly  the  same  way  that  electric  cir- 
cuits are  treated.  Ohm's  law  and  KirchhofTs  laws  can  be 
applied  to  the  magnetic  circuit  as  well  as  to  the  electric 
circuit.  In  cases  where  the  flux  paths  are  in  doubt  because 
of  the  introduction  of  large  air  gaps  into  circuits  otherwise 
of  iron,  approximate  methods  are  used  which  depend  upon 
estimations  of  the  paths  and  the  consequent  reluctances. 
In  view  of  the  fact  that  such  cases  cannot  be  solved  accu- 
rately by  any  known  means,  this  method  is  much  preferred 
to  other  and  more  complicated  methods  of  analysis. 

Prob.  27-7.     Map  the  field  as  in  Fig.  126,  about  two  iron 
bars  in  the  same  plane  one  inch  in  width,  one-quarter  inch  in 


THE  MAGNETIC  FIELD  241 

thickness  and  ten  inches  long,  whose  ends  are  separated  by  one 
inch.  Calculate  the  reluctance  of  a  filament,  also  the  total 
reluctance. 

Prob.  28-7.     What  would  be  the  approximate  reluctance  of 
an  air  gap  of  J  inch  between  the  two  bars  of  Prob.  27-7? 


SUMMARY    OF   CHAPTER  VII 

THE  MAGNETOMOTIVE  FORCE,  J,  is  the  LINE  IN- 
TEGRAL of  the  magnetizing  force,  H,  about  a  closed  circuit 
and  equals  0.47r  times  the  number  of  ampere-turns  linking  the 
circuit.  As  an  equation,  this  may  be  expressed 

Hdl  =  0.47rNI. 


TO  DEFINE  A  LINE  INTEGRAL,  consider  a  point  acted 
upon  by  a  variable  force  and  moving  along  a  path,  and  in- 
tegrate over  the  entire  length  of  the  path  the  products  ob- 
tained by  multiplying  the  force  at  every  point  in  the  path  by 
the  cosine  of  the  angle  between  the  direction  of  force  and  the 
path.  The  result  is  really  the  energy  expended  on  the  point 
by  the  force  in  traveling  the  length  of  the  path. 

THE  MAGNETIZING  FORCE  at  a  point  outside  a  long 
wire  carrying  a  current  is  expressed  by  the  equation 


0.21 
H  =— , 


where 


H  =  the  magnetizing  force  in  GILBERTS  PER  CENTI- 
METER, 

I  =  the  current  in  AMPERES, 

r  =  the  distance  of  the  point  from   the   center   of   the 
wire  in  CENTIMETERS. 

THE  FLUX  DENSITY  at  a  point  in  a  material  of  a  per- 
meability n  completely  surrounding  a  long  wire  which  is 
carrying  a  current  may  be  expressed  by 

0.2MI 

D     —  • 

r 

IN  THE  VICINITY  OF  A  SHORT  WIRE  carrying  a  current 
the  magnetizing  force  is 

H  =  -  (sin  «i  ±  sin  «*), 
243 


THE  MAGNETIC  FIELD  243 


where 


H  =  the  magnetizing  force  in  GILBERTS  PER  CENTI- 

METER, 

I  =  the  current  in  ABAMPERES, 
r  =  the  distance  of   the  point  from  the  wire  in  CEN- 

TIMETERS, 

ai  and  «2  are  angles  between  the  perpendicular  from  the 
point  to  the  wire  and  lines  drawn  from  the  point 
to  the  ends  of  the  wire. 

THE  TOTAL  FLUX  BETWEEN  THE  TWO  PARALLEL 
WIRES  (line  and  return)  of  a  transmission  line  can  be  found 
from  the  formula 

<j>  =  4  I  loge  --  maxwells  per  centimeter  of  line, 

where 

I  =  the  current  in  ABAMPERES, 
D  =  the  distance  between  the  centers  of   the  wires  in 

CENTIMETERS, 
r  =  the  radius  of  the  wires  in  CENTIMETERS. 

THE  FIELD  AT  A  POINT  WITHIN  A  CIRCULAR  CON- 
DUCTOR equals 

24  a 
:Vr' 
where 

I  =  the  current  in  ABAMPERES, 
M  =  the  permeability  of  the  material  of  the  conductor, 
r  =  the  radius  of  the  conductor  in  CENTIMETERS, 
a  =  the  distance  of  the  point   from   the  center  of   the 
conductor  in  CENTIMETERS. 

AT  THE  CENTER  OF  A  CIRCULAR  COIL  OF  A  SINGLE 
TURN  the  magnetizing  force  equals 


AT  THE  CENTER  OF  AN  ARC  OF  ONE  UNIT  IN  LENGTH 
and  with  a  radius  of  one  unit 


244      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

UNIT  CURRENT  can  thus  be  defined  as  that  current  which 
flowing  in  a  unit  length  of  an  electric  circuit  bent  into  an  arc 
of  unit  radius  will  produce  unit  magnetizing  force  at  the  cen- 
ter of  the  arc. 

ON  THE  AXIS  OF  A  CIRCULAR  COIL  OF  ONE  TURN 
the  magnetizing  force  equals 


H  =  — Esin30, 


where 


6  =  the  angle  between  the  axis  and  a  straight  line  joining 
the  point  and  the  wire. 

AT  ANY  POINT  ON  THE  AXIS  OF  A  SOLENOID 

27TNI 

H  =  — j —  (coson  —  cos«2) , 

where 

ai  and  «2  are  the  angles  between  the  axis  and  lines  join- 
ing the  point  to  each  end  turn. 

AT  THE  CENTER  OF  A  SOLENOID 

47TNI 


AT  ANY  POINT  ON  THE  MID-SECTION  OF  A  LONG 
SOLENOID 

4.NI 

~T 

THE    TOTAL    FLUX    THROUGH    THE    MID-SECTION 
OF  A  LONG  SOLENOID  with  an  air  core  equals 

47TNI 


<t>  =  BA 


A  long  air-core  solenoid  acts  as  if  practically  all  the  re- 
luctance were  within  the  coil  and  equal  to  —  •     The  fact  that  the 

£% 

flux  density  at  the  center  of  a  long  coil  is  uniformly  distri- 
buted over  the  cross-section  is  used  in  calibrating  a  ballistic 
galvanometer  to  be  used  to  measure  magnetic  flux. 


THE  MAGNETIC  FIELD  245 

THE  EXACT  EQUATION  FOR  THE  FLUX  IN  A  TOROID 
of  constant  permeability  equals 


where 

r  =  the  radius  of  the  toroid  in  CENTIMETERS, 
p  =  the  radius   of  the   stock   of  the   toroid   in   CENTI- 
METERS. 

THE  FLUX  SET  UP  BY  CONDUCTORS  IMBEDDED  IN 
IRON  can  be  found  approximately  by  plotting  the  approximate 
path  in  the  air,  computing  the  reluctance  of  this  path  and 
applying  the  line-integral  law  for  the  magnetic  circuit.  IN 
USING  THE  LINE-INTEGRAL  METHOD  for  circuits  of 
non-uniform  permeability,  great  care  must  be  exercised  always 
to  include  the  magnetic  whole  in  the  calculations.  The  value 
of  H  computed  for  a  point  in  the  air  will  not  hold  for  iron  placed 
at  that  point  of  the  circuit. 


PROBLEMS   ON   CHAPTER   VII 

Prob.  29-7.  A  solenoid  for  use  in  connection  with  flux 
measurements  has  a  primary  (air-core)  coil  7  centimeters  in 
diameter  and  75  centimeters  in  length  wound  with  525  turns 
of  cotton-covered  copper  wire,  which  can  safely  carry  3 
amperes.  What  will  be  the  magnetizing  force  in  gilberts 
and  what  flux  density  will  be  produced  at  a  point  at  the 
center  of  the  axis  of  the  coil  if  this  current  is  flowing? 

Prob.  30-7.  A  Leeds  and  Northrup  wall-type  galvanometer 
(Rg  =  117  ohms)  was  found  by  experiment  to  have  the  fol- 
lowing calibration  as  a  flux  meter: 

<f>N  =  d(4.625fl  +  868.75), 
where 

<f>N  =  flux  linkages  in  the  galvanometer  circuit, 
d  =  galvanometer  deflection, 
R  =  total  resistance  in  the  galvanometer  circuit. 

A  search  coil  of  100  turns  of  No.  36  wire  (mean  length  of  turn 
1.1  inches)  was  placed  so  that  it  linked  all  the  flux  lines  of  a 
standard  bipolar  telephone-receiver  magnet.  The  coil  was 
suddenly  moved  to  the  point  where  it  had  cut  all  the  flux  and 
the  deflection  was  30.35  centimeters.  What  was  the  flux  in 
maxwells  in  the  receiver  magnet? 

Prob.  31-7.  A  test  solenoid  used  in  connection  with  a  wall 
galvanometer  and  made  by  Robert  W.  Paul  of  London  has  a 
length  of  20  inches.  This  coil  is  so  wound  that  the  magneto- 
motive force  in  gilberts  per  centimeter  is  four  times  the  current 
carried  by  the  solenoid  winding.  The  secondary  coil  is  wound 
in  concentrated  form  around  the  solenoid  and  at  the  midpoint 
of  the  length.  What  must  be  the  number  of  turns  on  the 
solenoid  and  what  will  be  the  flux  linkages  of  the  secondary 
coil  neglecting  leakage? 

Prob.  32-7.  From  an  analysis  of  a  telephone  receiver, 
(Research  Bulletin  II,  Mass.  Inst.  Tech.),  the  equivalent 
magnetomotive  force  of  the  permanent  magnet  was  found  to 
be  181  gilberts.  The  average  useful  polar  flux  density 

246 


THE  MAGNETIC  FIELD  247 

produced  by  this  magnetizing  force  is  1369  gausses.  Find  the 
equivalent  reluctance  of  the  circuit.  On  the  spool  of  the 
receiver  magnets  are  wound  approximately  1300  turns  of  wire. 
What  increase  in  flux  would  a  current  of  2.56  milliamperes 
produce  assuming  the  permeability  to  remain  constant  through 
the  small  change?  The  dimensions  of  the  polar  surfaces  of 
this  bipolar  receiver  are  each  1.14  X  0.199  centimeters. 

Prob.  33-7.  Find  the  flux  density  at  the  polar  surfaces  of 
the  receiver  of  Prob.  32-7  when  the  additional  magnetomotive 
force  due  to  the  current  in  Prob.  32-7  acts.  (Assume  constant 
permeability.) 

Prob.  34-7.  A  cast-iron  toroid  of  circular  section,  (see 
Fig.  Ill),  T  =  6  inches,  n  —  r2  =  1.5  inches,  has  cut  in  it  a 
radial  slot  by  means  of  a  hacksaw.  The  width  of  the  slot  is  0.12 
inch.  If  there  are  410  ampere-turns  on  the  core,  what  will  be 
the  total  flux  in  the  core?  Plot  the  flux  density  in  the  core  in  a 
circumferential  direction  from  r\  to  r2,  neglecting  fringing.  Plot 
the  flux  with  the  same  ampere-turns  but  without  the  air  gap. 
What  will  be  the  difference  in  determining  the  total  flux  by 
the  exact  and  by  the  approximate  formulas? 

Prob.  35-7.  In  the  equation  for  the  flux  density  at  any 
point  in  the  toroid 

2NI 


(see  page  219)  why  is  it  not  accurate  to  find  an  average  flux 
density  by  finding  the  flux  at  a;  =  n  and  x  =  r2  and  taking  the 
mean? 

Prob.  36-7.  The  telephone  line  of  Fig.  93  is  to  be  strung 
with  the  same  spacing  between  x  and  y  on  the  same  poles  as 
the  power  line.  Compute  the  maximum  flux  linking  per  mile 
of  the  telephone  line  if  strung  as  follows: 

(a)  In  a  vertical  plane  passing  through  the  center  of  a  line 
joining  A  and  B,  and  perpendicular  to  it,  with  the  top  wire 
10  feet  below  the  line  joining  A  and  B; 

(6)  In  a  horizontal  plane,  with  the  center  of  a  line  joining 
x  and  y  directly  beneath  the  center  of  the  line  joining  A  and 
B,  and  10  feet  distant. 

Prob.  37-7. 

(a)  In  a  solenoid  100  centimeters  long  and  having  an  average 
diameter  of  20  centimeters,  what  percent  error  is  made  in 


248      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


computing  the  flux  density  at  the  center  by  the  use  of  the 
formula 


(6)  If  the  solenoid  were  40  centimeters  in  diameter,  what 
percentage  error  would  have  been  made? 

Prob.  38-7.  Derive  the  equation  for  the  magnetizing  force 
at  any  point  on  the  axis  of  a  solenoid  but  outside  of  the  solen- 
oid. 

'  Prob.  39-7.  An  air-core  solenoid  100  centimeters  long  and 
of  15  centimeters  average  diameter  is  wound  with  1200  turns, 
carrying  4  amperes.  Compute 

(a)  The  field  intensity  at  the  center; 
(6)  The  flux  density  at  the  ends; 

(c)  The  internal  reluctance  ; 

(d)  The  external  reluctance. 

Prob.  40-7.     From  the  data  in  Prob.  39-7,  compute 

(a)  The  internal  reluctance  drop; 

(6)  The  magnetic  difference  of  potential  between  the  ends; 

(c)  The  external  reluctance  drop. 

What  is  the  flux  density  at  a  point  outside  the  coil  on  the  axis 
of  the  solenoid  in  Prob.  39-7,  20  centimeters  from  the  nearer 
end? 


FIG.  128.     A  conductor  P  placed  at  the  bottom  of  a  slot. 

Prob.  41-7.     Derive  a  formula  for  computing  the  flux  set 
up  by  a  current  of  /  abamperes  in  the  conductor  P  in  Fig.  128; 
(a)  Leaving  the  bottom  of  the  slots  U  and  V; 
(&)  Leaving  the  ends  of  the  teeth  X  and  Y. 

Prob.  42-7.     In  Fig.  128, 

a  =0.82  inches 

b  =  0.90 

c  =  0.82       " 

d  =  2.80 

h  =  1.25 

k  =  0.075     " 


THE  MAGNETIC  FIELD  249 

Find  the  total  flux  set  up 

(a)  In  the  slot  in  which  the  conductor  lies; 

(6)  In  the  air  at  the  ends  of  the  teeth  R  and  S; 

(c)  In  the  slots  u  and  v; 

(d)  In  the  air  at  the  ends  of  the  teeth  x  and  y. 


CHAPTER  VIII 
INDUCED  VOLTAGES 

We  have  seen  in  previous  chapters  that  when  flux  is  varying 
through  a  coil,  there  is  a  voltage  produced  in  the  coil.  In 
this  chapter  we  will  consider  some  of  the  immediate  laws 
governing  voltages  and  currents  induced  in  this  way. 

79.  Change  of  Linkages.  Lenz's  Law.  The  general  law 
may  be  stated  as  follows.  Whenever  the  flux  linkages  with 
a  coil  are  changing,  there  is  a  voltage  induced  in  the  coil. 

We  have  learned  that  this  induced  voltage  is  proportional 
to  the  rate  of  change  of  the  flux.  In  fact,  we  defined  the 
maxwell  as  that  amount  of  flux  which  would  produce  one 
abvolt-second  impulse.  If  the  flux  through  a  coil  changes 
by  one'  maxwell,  the  time  for  the  change  occupying  one 
second,  there  will  be  produced  in  the  coil  one  abvolt. 

The  voltage  produced  in  a  coil  is  thus  equal  to  the  rate  of 
change  of  flux  linkages  through  it;  that  is, 

E  =  tf^abvolts,  (1) 

where 

E  =  the  voltage  produced  in  abvolts, 
N  =  the  number  of  turns, 
0  =  the  flux  in  maxwells. 

If  E  is  to  be  expressed  in  volts,  we  must  introduce  the 
factor  10~8  to  convert  from  the  c.g.s.  units  to  the  practical 
system,  thus: 

E  =N^  10-8  volts.  (2) 

The  direction  of  the  induced  voltage  may  be  determined 
by  the  following  law:  an  induced  voltage  is  always  in  such 

250 


INDUCED    VOLTAGES 


251 


a  direction  that  it  tends  to  produce  a  current  which  will 
oppose  the  change  of  flux. 

This  was  stated  by  Lenz  as  follows  and  is  called  Lenz's 
Law. 

Whenever  there  is  a  change  in  the  amount  of  magnetic 
flux  linking  an  electric  circuit,  voltage  is  set  up  tending  to 
produce  a  -current  in  such  a  direction  as  to  oppose  the 
change  in  flux. 

The  voltage  thus  set  up  is  directly  proportional  to  the  rate 
of  change  of  flux  linkages. 

In  Fig.  129,  which  shows  a  transformer,  suppose  that 
the  primary  current  is  in  the  direction  shown  by  the 
arrow,  thus  producing  a  flux 
in  the  core  in  the  direction 
shown  by  the  dotted  arrow, 
and  suppose  that  this  flux 
is  decreasing.  There  will 
be  produced  in  the  second- 
ary coil  a  voltage  in  the 
direction  of  the  arrow 
shown  on  the  secondary 
winding,  that  is,  in  such  a  direction  that  a  current  flowing 
in  this  direction  would  increase  the  flux,  or  in  other  words, 
in  a  direction  such  as  to  tend  to  prevent  any  decrease  of 
flux.  Conversely,  if  the  primary  current  is  increasing,  the 
secondary  voltage  will  be  in  the  reverse  direction,  and  hence 
tend  to  oppose  an  increase  of  flux. 

Since  the  induced  voltage  is  proportional  to  the  rate  of 
change  of  flux,  a  very  high  voltage  may  be  obtained  by 
causing  the  flux  to  change  rapidly.  This  principle  is  made 
use  of  in  an  induction  coil  such  as  is  shown  in  Fig.  130. 
An  iron  core  is  wound  with  a  heavy  winding  capable  of 
carrying  a  large  current  and  thus  magnetizing  it  strongly. 
This  winding  is  called  the  primary.  The  core  is  usually 
made  open,  rather  than  a  closed  ring,  for  reasons  which  will 
appear  more  fully  in  the  next  chapter.  The  main  reason 


FIG.  129. 


Diagram  of  a  trans- 
former. 


252      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


may  be  briefly  stated  to  be  in  order  that  the  flux  may  collapse 
rapidly.  It  is  also  made  of  fine  iron  wire  to  reduce  eddy 
currents.  Over  this  primary  winding  is  wound  the  second- 
ary of  a  large  number  of  turns. 
The  primary  circuit  flows 
through  a  vibrator,  which 
operates  in  the  same  manner 
as  the  clapper  of  an  electric 
bell.  When  the  current  is 
interrupted  by  this  vibrator, 
the  flux  decreases  rapidly 


FIG.  130. 


Diagram  of  an  induc- 
tion coil. 


from  a  maximum  value  to 
nearly  zero.  The  rapid 
change  of  flux  linking  the  secondary  produces  a  high  voltage 
in  the  secondary. 

As  an  example,  suppose  that  the  core  is  4  square  centi- 
meters in  cross-section  and  is  magnetized  to  a  maximum  of 
6000  gausses,  thus  producing  a  total  flux  of  24,000  maxwells. 
Assume  that  the  vibrator  reduces  this  flux  to  zero  in  one- 
one-thousandth  of  a  second.  Let  there  be  20,000  second- 
ary turns.  The  voltage  produced  in  the  secondary  will 
then  be 


(3) 


Induction  coils  can  be  produced  in  this  manner  which  will 
induce  voltages  of  several  hundred  thousand  volts.  The 
principal  commercial  use  of  the  induction  coil  at  present  is 
in  the  ignition  system  of  automobiles.  The  vibrator  is  here 
replaced  by  an  interrupter  which  is  usually  mechanically 
driven  by  the  engine.  The  adjustment  of  this  interrupter 
contact  determines  the  period  in  the  cycle  at  which  the  cir- 
cuit is  opened  and  thus  the  point  at  which  the  spark  takes 
place.  The  primary  current  is  obtained  from  the  storage 
battery  of  the  car. 

A  condenser  is  placed  across  the  interrupter  to  reduce 


E  =  20,000  X  ^£r  X  10-8  =  4800  volts. 


INDUCED    VOLTAGES  253 

sparking  at  the  contacts.     The  size  of  this  condenser  is 
chosen  to  give  maximum  rate  of  decrease  of  flux. 

Much  more  important  from  a  commercial  standpoint  is 
the  transformer,  which  operates  on  the  same  principle.  A 
primary  winding  of  heavy  wire,  carrying  a  large  current  at 
low  voltage,  produces  a  large  flux  in  a  core,  which  in  turn 
produces  a  high  voltage  in  a  secondary  winding.  This  is 
the  step-up  transformer,  which  transforms  from  a  low  volt- 
age to  a  higher  voltage.  Conversely,  we  may  have  a  pri- 
mary of  a  large  number  of  turns  carrying  a  small  current 
which  produces  a  flux,  giving  a  heavy  current  at  low  voltage 
in  a  secondary  winding  of  few  turns  of  heavy  wire.  This  is 
the  step-down  transformer.  The  voltage  ratio  of  the  trans- 
former is  approximately  the  same  as  the  ratio  of  the  number 
of  turns  on  the  primary  and  secondary. 

In  the  transformer  the  flux  does  not  vary  abruptly,  as  it 
does  in  the  induction  coil, 
but  changes  smoothly  and 
periodically.     In  fact,  alter- 
nating currents  are  usually     t  /  \  / 

,  .  •  i    i  i         §  /  \  Time  Axis/ 

very  nearly  sinusoidal  and     o  • ; V 

the  flux  in   the   core  of  a     s  i      \ 


J-Sec: 

transformer   is  also  nearly 

a    sinusoid.      This    means 

that  it  varies  harmonically 

with   the   time.     In   other 

words,  if  we  plot  the  flux 

density  in  the  core  against  the  time,   we  shall  obtain  a 

sine  curve,  for  instance  Fig.  131,  the  equation  of  which  is 

B  =  £maxsin27r/J.  (4) 

The  symbol  /  is  the  frequency,  which  in  American  practice 
is  usually  60  cycles  per  second.  The  voltage  produced  in 
the  secondary  will  be 

E  =  N  10~8  1  (A B^  sin  2x/O,  (5) 


254      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


where 


N  =  the  number  of  secondary  turns, 

A  —  the  cross-sectional  area  of  the  core. 


This  becomes  upon  differentiating 

E  =  27T  NA  Bmax/cos  2<irfi  ,  (6) 

which  it  will  be  noted  contains  a  cosine  instead  of  a  sine. 

This  equation  is  plotted  in  Fig.  132.  Note  that  the  volt- 
age is  at  its  maximum  when 
the  flux  is  zero,  that  is,  when 
the  flux  is  changing  at  its 
maximum  rate.  Note  also 
that  when  the  flux  is  at  max- 
imum it  is  changing  at  zero 
rate,  and  the  secondary  volt- 
age is  accordingly  at  that 
time  zero.  We  say  that  the 
voltage  in  such  a  case  is  a 


Time 


Axis 


FIG.  132.  When  flux  varies  ac- 
cording to  the  sine  curve  of 
Fig.  131,  the  induced  voltage 
varies  according  to  the  cosine 
curve  shown  here. 


quarter  of  a   period  out  of 
phase    with    the    flux,    that 
is,  in  quadrature  with  the  flux. 

The  complete  study  of  the  action  of  a  transformer  is  an 
interesting  problem,  but  it  must  be  postponed  until  the 
laws  of  the  electric  and  magnetic  circuits  are  more  thor- 
oughly understood. 

Prob.  1-8.  The  strength  of  the  earth's  magnetic  field  at  a 
given  point  is  0.6  gauss  and  the  angle  of  dip  is  70°  with  the 
horizontal.  If  a  coil  of  900  turns  is  held  vertically  with  its 
plane  perpendicular  to  a  north-and-south  line  and  then  rotated 
about  its  vertical  axis  through  90°  in  0.02  second,  what  average 
voltage  will  be  produced  in  the  coil  by  this  movement?  What 
would  be  the  average  voltage  if  the  coil  were  moved  through 
only  30°  at  the  same  speed?  Through  50°?  Area  =  A  sq.  cm. 

Prob.  2-8.  If  the  primary  coil  of  Fig.  130  is  wound  with 
200  turns  of  wire  to  produce  a  total  flux  of  24,000  maxwells, 
what  average  voltage  will  appear  at  the  points  of  the  vibrator 
when  the  circuit  is  opened  if  the  flux  dies  down  to  zero  in  0.001 
second? 


INDUCED    VOLTAGES  255 

Prob.  3-8.  If  in  a  transformer  core  Bm  =  70,000,  plot  the 
flux  density  with  time  for  one  cycle  where  /  =  60,  assuming  a 
sinusoidal  variation. 

Prob.  4-8.  If  the  cross-section  of  the  core  of  the  transformer 
of  Prob.  3-8  is  1.5  square  inches,  plot  E  (the  secondary  voltage) 
against  time  for  one  cycle,  where  N  =  2500  (secondary  turns). 
Use  the  same  scale  for  time  as  in  Prob.  3-8  and  superimpose 
this  plot  on  the  plot  of  Prob.  3-8. 

80.  Self-induction.  Coefficient.  A  voltage  is  induced  in 
a  coil  whenever  the  number  of  flux  linkages  is  changing. 
This  is  true  whether  there  is  a  current  in  the  coil  or  not. 
If  we  have  only  a  single  coil,  which  is  carrying  a  current  and  is 
thus  producing  a  flux,  it  will  have  a  voltage  produced  in  it 
whenever  this  flux  (which  it  is  itself  producing)  is  changing 
in  amount.  Such  a  voltage  we  call  a  voltage  of  self-induc- 
tion. 

We  have  seen  that  an  induced  voltage  is  always  in  such 
a  direction  as  to  tend  to  oppose  the  cause  that  produces  it. 
A  voltage  of  self-induction  is  therefore  always  in  such  a 
direction  as  to  oppose  a  change  in  current.  If  the  current 
is  building  up  in  a  coil,  that  is,  increasing,  the  self-induced 
voltage  will  be  opposite  to  the  current  in  direction.  If 
the  current  in  a  coil  is  decreasing,  the  self -induced  voltage 
will  be  in  the  same  direction  as  the  current. 

The  number  of  flux  linkages  which  are  set  up  by  a  coil 
with  its  own  turns,  when  the  coil  is  carrying  unit  current, 
is  called  the  coefficient  of  self-induction  of  the  coil.  The 
c.g.s.  unit  of  the  coefficient  of  self-induction  is  therefore 
the  number  of  linkages  per  abampere  in  the  coil,  and  is 
called  the  abhenry.  The  practical  unit  is  10~9  times  as 
large,  and  is  called  the  henry.  A  coil  accordingly  has  a 
coefficient  of  self-induction,  or  more  briefly,  simply  an 
inductance,  of  one  henry  when  a  current  of  one  ampere 
flowing  in  the  coil  will  produce  108  linkages.  That  is, 

L  =  -       10-8  henries,  (7) 


256      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

where 

L  =  the  inductance  in  the  coil  in  henries, 
N  =  the  number  of  turns, 

0  =  the  total  flux  produced  by  a  current  of  7  am- 
peres flowing  in  the  coil. 

This  assumes  that  all  of  the  flux  links  every  turn.  If  the 
flux  does  not  link  all  of  the  turns,  we  should,  of  course, 
write  the  actual  number  of  linkages  produced  instead  of 
the  product  N<l>. 

The  inductance  is  a  constant  which  depends  only  upon 
the  dimensions  of  a  coil.  No  iron  is  used  in  the  construction 
of  the  coil.  The  flux  is  then  proportional  to  the  current  and 
L  will  have  the  same  value  no  matter  what  number  of 
amperes  is  used  in  finding 


The  same  is  not  true  when  an  iron  core  is  used.  Then  the 
flux  is  not  strictly  proportional  to  the  current,  although 
for  small  magnetizing  forces  it  may  be  nearly  enough  pro- 
portional so  that  the  formulas  may  still  be  used  with  en- 
gineering accuracy. 

In  speaking  of  the  inductance  of  a  coil  or  a  circuit  we  mean 
at  present  a  coil  or  a  circuit  in  which  no  iron  is  present  or  in 
which  the  flux  density  in  the  iron  used  is  small  and  hence 
the  flux  is  proportional  to  the  current.  In  the  ordinary 
coil  with  an  iron  core  at  high  flux  densities,  the  flux  is  not 
at  all  proportional  to  the  current,  and  the  inductance  of  the 
coil  is  not  a  constant,  but  depends  upon  the  value  of  cur- 
rent used  in  measuring  it.  In  such  a  case  the  term  "  coeffi- 
cient of  self-induction  "  ceases  to  have  a  definite  meaning. 
We  will  deal  with  such  cases  in  the  next  chapter. 

Suppose  that  we  have  a  coil  not  containing  iron  and 
having  an  inductance  of  one  henry.  Assume  that  this 
coil  is  carrying  a  current  which  is  increasing  uniformly  at 


INDUCED    VOLTAGES  257 

the  rate  of  one  ampere  per  second.  Then,  since  in  such  a 
coil  one  ampere  produces  108  linkages,  the  number  of  link- 
ages will  be  increasing  at  the  rate  of  108  per  second.  We 
have  seen,  however,  that  when  the  linkages  in  a  coil  are 
changing  at  the  rate  of  108  per  second,  there  is  an  induced 
voltage  in  the  coil  of  one  volt.  We  are  thus  led  to  another 
definition  of  the  henry. 

A  coil  has  an  inductance  of  one  henry  if  there  is  induced  in  it 
one  volt  where  the  current  is  changing  at  the  rate  of  one 
ampere  per  second. 

This  definition  may  be  extended  and  expressed  as  follows. 
The  voltage  induced  in  a  coil  by  self-induction  is  equal  to 
the  inductance  times  the  rate  of  change  of  current.  In  prac- 
tical units,  the  voltage  induced  in  a  coil  by  self-induction  is 
equal  to  the  inductance  in  henries  times  the  rate  of  change  of 
current  in  amperes  per  second;  that  is, 


The  minus  sign  is  introduced  to  indicate  that  the  voltage 
is  in  a  direction  opposite  to  the  current  when  the  current  is 
increasing.  This  we  have  seen  to  be  the  case,  since  the 
voltage  is  always  in  such  a  direction  as  to  oppose  the  cur- 
rent change. 

This  very  simple  formula,  "The  induced  voltage  is  equal 
to  the  inductance  times  the  rate  of  change  of  current,"  will  be 
found  of  great  importance  in  all  electrical  theory. 

Prob.  5-8.  What  is  the  inductance  in  the  primary  coil  in 
Prob.  2-8  if  the  current  in  the  primary  is  1.7  amperes  and  the 
flux  has  a  value  of  24,000  maxwells? 

Prob.  6-8.  What  voltage  is  induced  in  a  coil  having  0.002 
henry  inductance  when  the  current  is  changing  at  the  rate  of 
5  X  104  amperes  per  second? 

Prob.  7-8.  A  total  voltage  of  3000  is  induced  at  a  certain 
instant  in  a  coil  of  2500  turns  of  wire.  At  what  rate  is  the 
current  changing  if  L  =  0.32  henry? 


258      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  8-8.  A  coil  of  350  concentrated  turns  of  wire  was 
tested  for  flux  with  a  ballistic  galvanometer.  It  was  found  that 
the  coil  was  linked  by  a  flux  of  110,000  maxwells  when  3  am- 
peres flowed  in  the  winding.  Find  L. 

81.  Transients  in  Inductive  Circuits.  We  have  seen  that 
when  a  current  is  varying  in  a  coil,  the  change  of  linkages 
produced  causes  a  voltage  to  be  set  up  in  the  coil  of  value 

di 
~Ldt 

and  this  voltage  of  self-induction  is  opposed  to  the  change 
of  current. 

When  a  coil  is  switched  on  a  circuit  and  the  current  starts 
to  build  up  through  the  coil,  this  voltage  of  self-induction 
opposes  this  increase  of  current,  and  therefore  causes  it  to 
build  up  slowly.  We  can  readily  analyse  the  way  in  which 
this  takes  place. 

Let  us  consider  the  simple  circuit  of  Fig.  133.  A  coil  of 
L  henries  inductance  is  connected  in  series  with  a  resistance 

of  R  ohms  and  a  battery  giving 
a  constant  voltage  of  E  volts. 
The  circuit  is  closed  by  a  switch. 
We  will  assume  that  the  resist- 
ance R  includes  all  the  resistance 
of  the  circuit,  that  is,  the  value 
taken  for  R  will  not  only  be  the 
FIG.  133.     A  circuit  contain-     external   resistance  but  also  the 
ing  an  inductance  L,  a  re-          ,  .    ,,  .  ,  .    ,, 

sistanceBandane.m.f.£.     vajue    of    the    resistance   of  the 

coil  itself  and  of  the  battery. 

We  know  by  Kirchhoff  s  laws  that  the  sum  of  the  volt- 
ages around  this  circuit  is  zero.  It  is,  in  fact,  equal  to  zero 
at  every  instant,  provided  all  of  the  voltages  are  taken  into 
account.  When  we  are  considering  a  current  which  is 
changing  in  value,  there  is,  in  addition  to  the  battery  volt- 
age and  resistance  drop,  a  voltage  of  self-inductance  which 
must  be  included  if  our  equation  is  to  give  a  correct  value. 


INDUCED    VOLTAGES  259 

Accordingly  when  the  current  is  building  up  in  a  coil  we 
have  three  voltages  to  consider,  —  first,  the  voltage  of  the 
battery,  second,  the  resistance  drop,  and  third,  the  voltage 
of  self  -inductance.  The  two  latter  are  opposed  to  the  volt- 
age of  the  battery.  Expressing  these  facts  mathemati- 
cally gives  us  the  differential  equation 

E  -  Ri  -  L^  =  0.  (9) 

This  equation  expresses  the  conditions  for  any  instant  in  the 
circuit. 

If  we  solve  this  differential  equation,  we  shall  obtain  an 
expression  for  the  current  in  terms  of  the  time,  that  is,  an 
equation  which  will  show  the  manner  in  which  the  current 
will  vary  after  closing  the  switch. 

Transposing  and  simplifying,  we  have 

(E  -  Ri)dt  =  L  di,  (10) 

which  reduces  to 


The  variables  are  now  separated  and  we  may  integrate 
both  sides  of  the  equation,  giving 


where  C  is  the  constant  of  integration.     This  equation  may 
be  put  into  the  form 

+c 

.  (13) 


The  logarithm  above  is,  of  course,  to  the  base  e. 

To  determine  the  constant  of  integration,  we  utilize  the 
known  fact  that  at  the  instant  of  closing  the  switch  the 


260      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


current  in  the  circuit  is  equal  to  zero. 

equation  the  pate  of  values 

t  =  0 
i-O, 


we  thus  obtain 


c 


Inserting  into  our 


(14) 


(15) 


Inserting  this  value  as  the  constant  of  integration  in  the 
equation  and  transposing,  we  have  as  a  final  result 

HI  Ri 

i  =  |  (1  -  *~T).  (16) 

This  equation  gives  the  manner  in  which  the  current  varies 
after  closing  the  switch.  It  is  plotted  in  Fig.  134.  The 
curve  shows  that  the  current  increases  at  first  very  rapidly, 

Tjl 

then  more  slowly  and  finally  approaches  the  value  •=  •     In 

the  first  instant  after  the  switch  is  closed,  the  voltage  of 

self-inductance  prevents 
the  current  from  coming 
immediately  to  its  final 
value.  As  tune  goes  on, 
however,  the  current  in- 
creases more  slowly,  until 
it  is  practically  steady  and 
the  effect  of  the  induc- 
tance disappears.  The  cur- 
rent then  has  its  steady- 


FIG.  134.  When  the  switch  in  the 
circuit  of  Fig.  133  is  closed,  the 
current  rises  according  to  this 
curve. 


state  value  as  given  by 
Ohm's  law.  We  now  see 
why  it  has  been  stated  several  times  that  Ohm's  law  in  its 
simple  form  must  be  applied  only  when  matters  in  the 
electric  current  have  reached  their  steady  condition. 

Note  carefully  that  the  current  follows  the  curve  of  Fig. 
134,  only  when  E,  R  and  L  are  all  constant.  When  the 
circuit  links  a  magnetic  circuit  in  which  iron  is  present,  L  is 
not  a  constant  and  the  growth  of  the  current  takes  the  form 


INDUCED    VOLTAGES  261 

shown  in  Fig.  134a.  Since  equation  16  applies  only  to  a 
curve  of  the  form  shown  in  Fig.  134,  it  cannot  be  used  in 
connection  with  a  circuit  linked  by  iron. 


FIG.  134a.  An  oscillogram  of  the  growth  of  the  current  in  a  circuit 
linking  an  iron  circuit.  The  alternating  wave  at  the  bottom  is  a  60- 
cycle  wave  for  determining  the  time.  Taken  at  M.I.T.  by  Prof.  F.  S. 
Dellenbaugh  of  the  Research  Division  of  the  Electrical  Engineering  De- 
partment. 

From  the  above  results,  we  see  that  it  takes  time  to  set 
up  a  current  through  an  inductive  circuit.  The  effect  is  very 
similar  to  the  corresponding  one  in  mechanics.  It  takes 
time  to  set  a  heavy  car  in  motion.  We  say  that  the  car  has 
inertia,  and  that  a  force  must  be  applied  for  a  certain  period 
in  order  to  overcome  the  inertia  of  the  car.  In  the  same  way 
an  electromotive  force  must  be  applied  for  a  certain  period 
in  order  to  overcome  the  inductance  of  the  circuit.  The 
inductance  of  a  circuit  is  hence  electrical  inertia. 

If  after  a  car  has  once  been  set  going  we  remove  its  driving 
force,  it  will  coast  for  a  considerable  time  before  stopping. 
Similarly,  if  a  current  is  set  up  in  an  inductive  circuit  and 
the  electromotive  force  of  the  circuit  then  removed,  the 
current  will  persist  for  a  period  before  coming  to  rest  at 


262      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


zero.     This  effect  we  can  analyse  in  a  manner  similar  to 
that  already  used. 

Suppose  in  Fig.  135  that  the  battery  E  is  furnishing  a 
steady  current  I  to  the  inductance  L  and  the  resistance  R. 
Let  us  now  close  a  switch  which  short  circuits  the  battery 
and  cuts  it  out  of  circuit,  the  battery,  of  course,  being  pro- 
tected against  short  circuit  by  additional  resistance.  The 
_  ^  current  through  the  induct- 

ance will  not  instantly  stop 
but  will  die  out  gradually. 
There  is  no  e.  m.  f.  in  the 
right-hand  loop  of  this 
circuit  after  the  switch  is 
closed  except  the  voltage 
of  self  inductance.  Apply- 
.  T^.  i  i  «.»  i 
?*  Kirchhoff's  law  to  the 

instantaneous  condition   in 


FIG.  135.  The  switch  cuts  out  the 
battery  from  the  circuit  of  L  and 
R  and  short  circuits  them. 

the  circuit,  we  obtain 


(17) 


which  can  be  reduced  to 

* 


-**• 


where  the  variables  are  now  separated.     Integrating,  we 

obtain  p/ 

log  t-  _-.«.'+  Ct(  (19) 

where  €2  is  a  constant  of  integration.     This  equation  may 
be  put  into  the  form  Rt 

i  =  €~L    e<\  (20) 

We  know  that  at  the  instant  of  closing  the  switch,  the  current 
has  the  value  I.    Accordingly  from  the  relations 


inserted  into  the  current  equation,  we  obtain 


<?', 


(22) 


INDUCED    VOLTAGES 


263 


and  inserting  this  value  for  the  constant  of  integration,  we 
obtain 

i  =  7e~?  (23) 

for  the  relation  between  the  current  and  the  time,  after  the 
switch  is  closed. 

This  equation  is  plotted  in  Fig.  136.  The  current  dies 
out  exponentially  and  approaches  zero  as  time  goes  on. 
The  manner  in  which  the  current  dies  out  in  a  circuit  is 
consequently  the  exact  reverse  of  that  in  which  the  current 
builds  up  when  the  voltage  is  applied. 

It  will  be  noted  that  mathematically  the  current  never 
equals  zero.  No  matter  what  value  of  t  we  use,  the  equation 
will  give  a  value  of  the 
current,  which  may  be, 
indeed,  exceedingly  small, 
but  which  still  exists.  The 
current  therefore  ap- 
zero  but  never 
it.  Practically, 
the  current  will 
short  interval  of 


preaches 

reaches 

however, 

within  a 

time  become  so  small  that 

it  may  be  entirely  neglected. 

There  are  always  voltages 

of    small   magnitude    present    in 

small     current     flowing     in     one 


FIG.  136.  The  current  in  the  circuit 
of  Fig.  135  decays  on  short  circuit 
according  to- this  curve. 


a  circuit  and  hence  a 
direction  or  another. 
These  small  voltages  may  be  due  to  thermal  or  chemical 
electromotive  forces,  or  some  similar  cause.  A  short  in- 
terval of  time  after  closing  the  switch  above,  the  current 
becomes  so  small  that  it  becomes  merged  with  these  minute 
currents.  To  all  intents  and  purposes,  it  is  then  zero.  It 
will  be  noted,  however,  that  the  time  that  it  takes  for  the 
current  to  reduce  to  zero  cannot  be  determined  experimen- 
tally. Neither  can  we  state  the  exact  length  of  time  that 
it  will  take  for  the  current  to  reach  its  final  value  when  the 


264      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

voltage  is  first  applied.  The  time  taken  for  it  to  reach  any 
given  fraction  of  its  final  value  is,  however,  a  perfectly  defi- 
nite matter. 

Prob.  9-8.  In  Fig.  133,  let  R  =  10  ohms,  L  =  0.0001  henry 
and  E  =  15  volts.  How  long  after  the  switch  is  closed  will  the 
current  rise  to  one-half  and  three-fourths  respectively  of  its 
final  value?  Plot  the  growth  of  the  current  by  finding  five 
points. 

Prob.  10-8.  If,  in  Prob.  9-8,  L  is  made  1.0  henry,  find  the 
time  as  before. 

Prob.  11-8.  If  the  circuit  of  Fig.  135  has  the  following  con- 
stants : 

R  =  50, 
L  =  0.5, 
E  =  100, 
r  =  12  (internal  resistance  of  battery), 

and  the  switch  is  closed  when  the  current  has  reached  its  final 
value,  at  what  rate  will  the  current  be  decreasing  when  t  = 
0.008  second?  What  will  be  the  time  necessary  for  the  current 
to  reduce  to  half  its  original  value?  What  is  the  initial  rate  of 
decrease? 

82.  Time  Constant.  The  greater  the  inductance  of  the 
circuit,  that  is,  the  ~more  inertia  it  has,  the  longer  it  will 
take  for  a  given  electromotive  force  to  set  up  a  given 
amount  of  current  through  it.  With  a  fixed  value  of  induc- 
tance, moreover,  the  greater  the  value  of  resistance  asso- 
ciated with  it,  the  less  will  be  the  final  value  of  current 
obtained,  and  accordingly  the  more  quickly  will  a  given 
electromotive  force  reach  any  given  fraction  of  this  final 
value.  The  addition  of  a  resistance  thus  masks  the  effect 
of  the  inductance.  The  less  inductance  a  circuit  has,  the 
more  quickly  a  current  can  be  built  up  through  it.  The 
more  resistance  it  has,  on  the  other  hand,  the  less  current 
can  be  passed  through  anyway;  and  so  the  less  will  the 
effect  of  the  inductance  be  felt. 

We  can  express  these  facts  exactly  by  saying  that  the  cir- 


INDUCED    VOLTAGES 


cuit  has  a  certain  time   constant. 
called  T,  and  has  the  value 


R 


265 
This  time  constant  is 

(24) 


The  time  constant  (in  seconds)  of  a  circuit  is  therefore  its 
inductance  in  henries  divided  by  its  resistance  in   ohms. 
If  a  circuit  has  an  induct- 
ance   of   2    henries    and  a 
resistance  of  1000  ohms,  its 
time  constant  will  be  0.002 
second. 

The  meaning  of  the  time 
constant  may  be  made  clear 
by  Fig.  137,  which  shows 
the  curve  for  the  decrease 
of  current  through  an  in- 
ductive circuit  after  the 
voltage  is  removed.  At 
the  end  of  the  time,  T,  after 
removing  the  electromotive 


to   decrease   at   the   rate  —f-  ,  it 


would  become  zero  in  ~  seconds. 
K 


n   is  the  time  constant  of   the 

K 


force,  the  current  will  have 

decreased  to  a  certain  frac-        circuit. 

tional  part  of  its   original 

value.     The  amount  of  the  current  at  'this  time  may  be 

found  by  inserting 

t  =  T  (25) 


or 


in  the  equation 

i  =  Ie~L 
for  the  current.     This  gives  evidently 

i  =  Ie~l 


**' 


(26) 

(27) 
(28) 


266      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


or 


i  =  - 

€ 


(29) 


and  since 


e  =  2.718, 


it  may  be  written  if  we  wish  (using  2.718  as  a  sufficiently 
precise  value  of  e) 

"  ~"~  T  (30) 


2.718 


J  =  0.3687. 


The  time  constant  of  an  inductive  circuit  is  hence  the  time 
in  which  the  current  will  decrease  to 


of  its  original  value  after  the  electromotive  force  of  the 
circuit  is  removed.  The  curve  for  the  increase  of  the  cur- 
rent, Fig.  138,  is  exactly 
the  same  as  the  curve  of 
Fig.  137,  except  that  it  is 
turned  upside  down.  We 
may  accordingly  state  the 
time  constant  of  a  circuit 
as  follows.  The  time  con- 

stant  of  a  circuit  is  the  time 

after      the      electromotive 

The  current  would  reach  force  of  the  circuit  is 
changed  for  the  current  in 
the  circuit  to  make  approxi- 
mately sixty-three  percent 
of  its  complete  change. 

For  a  current  starting  at  zero,  it  is  the  time  after  the  electro- 
motive force  is  applied  for  the  current  in  the  circuit  to  reach 
approximately  sixty-three  percent  of  its  final  value.  The 
exact  percentages  may  be  worked  out  to  as  great  an  ac- 
curacy as  desired  from  the  known  value  of  e. 


EI  T 

the  value  ^  in  ~  or  T  seconds  if 
K       K 

it  continued  to  grow  at  the  same 
rate  at  which  it  started. 


INDUCED    VOLTAGES  267 

The  time  constant,  however,  has  a  certain  further  sig- 
nificance which  should  be  understood. 

In  Fig.  137,  let  us  find  the  tangent  to  the  curve  when  t  is 
zero.  The  slope  of  this  tangent  is 

di 
dt' 
But 

Rt 

i  =  Ie~L 

because  this  is  the  equation  for  the  current  when  the  in- 
pressed  voltage  is  removed. 
Therefore,  differentiating, 

di          RI  -f 
dt  =     "Lf   L' 

When  t  is  zero,  this  equation  for  the  slope  reduces  to 

di  _       RI 
dt~     '~L' 

Extend  the  tangent  at  the  zero  point  until  it  cuts  the  time 
axis,  that  is,  the  axis  of  abscissas.  From  the  geometry  of 
the  triangle  of  the  figure,  it  will  cut  it  at  a  point  distant  from 
the  origin  an  amount 


that  is, 

'-•§•  (33) 

We  thus  see  that  the  tangent  to  the  curve  at  the  /-axis 
cuts  the  horizontal  axis  at  a  distance  from  the  origin  equal 
to  the  time  constant. 

The  tangent,  however,  gives  the  rate  at  which  the  current 
starts  to  decrease.  This  leads  us  to  the  following  statement. 

The  time  constant  of  an  inductive  circuit  is  the  time  in 


268      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

which  the  current  would  decrease  to  zero  after  the  electro- 
motive force  was  removed,  provided  it  continued  to  decrease 
at  its  initial  rate.  Similar  treatment  applies  to  Fig.  138. 
The  time  constant  may  also  be  stated  thus:  the  time  con- 
stant of  an  inductive  circuit  is  the  time  in  which  the  current 
would  reach  its  final  value  in  the  circuit  after  the  appli- 
cation of  an  electromotive  force  to  the  circuit,  provided  that 
it  continued  to  increase  at  the  rate  at  which  it  initially  started 
to  increase. 

When  the  switch  is  closed  in  an  inductive  circuit,  the  cur- 
rent changes  first  rapidly  and  then  more  gradually.  If  it 
did  not  taper  out,  but  continued  to  change  at  the  initial 
rate,  it  would  complete  the  entire  change  in  a  time  equal  to 
the  time  constant  of  the  circuit. 

As  an  illustration  of  these  principles,  suppose  that  the 
field  circuit  of  a  dynamo  consists  of  a  large  coil  of  wire  having 
a  resistance  of  ten  ohms  and  an  inductance  of  two  henries. 
The  time  constant  of  this  circuit  will  be 

T  =  ^  =  ^  =  0.2   second.  (34) 

K        iU 

Suppose  that  we  close  the  switch  which  supplies  an  elec- 
tromotive force  of  100  volts  to  this  circuit.  The  final  cur- 
rent will  be  -TTT-  or  10  amperes.  At  the  end  of  two-tenths 

of  a  second  after  closing  the  switch,  the  current  will  have 
reached  63  percent  of  this  final  value,  or  6.3  amperes. 
Upon  first  closing  the  switch,  the  current  will  increase  at  the 
rate  of  50  amperes  per  second,  that  is,  at  such  a  rate  that  if 
it  continued  at  this  rate  it  would  reach  its  final  value  of  10 
amperes  at  the  end  of  0.2  second.  Assume  that  we  wish  to 
know  the  current  in  this  circuit  after  a  lapse  of  one  second. 

The  value  will  be 

10x1 
i  =  10  (1  -  e — 27)  =  10  (1  -  e~6) 

=  10  (1  -  0.007)  (35) 

=  9.93  amperes; 


INDUCED    VOLTAGES  269 

that  is,  99.3  percent  of  its  final  value.  At  the  end  of  one 
second,  then,  the  current  has,  as  nearly  as  we  could  easily 
measure,  completely  reached  the  value  it  will  maintain 
continuously. 

Prob.  12-8.   In  Prob.  9-8,  what  is  the  value  of  the  current 

when  t  =  —  ? 
H 

Prob.  13-8.  If  a  constant  voltage  is  suddenly  applied  to  a 
circuit  containing  resistance  and  inductance,  what  is  the  initial 
rate  of  increase  of  the  current? 

Prob.  14-8.  If  the  rate  of  growth  of  current,  at  the  first  in- 
stant the  switch  is  closed  in  a  circuit  containing  inductance, 
resistance  and  a  battery  is  20  amperes  per  second,  and  the 
final  value  of  current  is  2  amperes,  what  is  the  time  constant? 

Prob.  15-8.  In  Fig.  133,  assuming  the  resistance  of  the 
battery  to  be  negligible  and  taking  the  following  constants: 

R  =  20, 
L  =  0.9, 

plot  the  growth  of  the  current  after  the  switch  is  closed.  Also 
draw  the  tangent  to  this  curve  at  t  =  0.  Show  on  the  plot 
where  the  tangent  intersects  the  straight  line  /  =  E/  R.  Com- 
pute the  time  constant  and  show  its  value  on  the  diagram. 
Do  the  same  for  the  decay  of  current  after  closing  the  short- 
circuiting  switch  in  Fig.  135,  assuming  the  same  constants. 

83.  Inertia  of  an  Electric  Circuit.  When  an  electric 
current  passes  through  an  electric  circuit,  there  is  a  move- 
ment of  electrons  through  the  metal.  It  takes  energy  to 
set  the  electrons  in  motion,  that  is,  they  have  inertia.  We 
have  seen  that  we  call  the  measure  of  this  inertia  reaction 
the  inductance  of  the  circuit. 

Similarly  the  electrons  when  moving  set  up  a  magnetic  field 
which  represents  kinetic  energy.  That  is,  there  is  energy 
stored  in  a  circuit  due  to  its  inductance.  In  order  to  set  up 
a  current  in  a  circuit,  it  is  accordingly  necessary  to  apply 
a  certain  amount  of  energy  to  be  stored  in  the  magnetic  field, 
in  addition  to  any  energy  losses  in  the  resistance.  We  will 


270      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

compute  the  amount  of  this  energy  storage  and  see  how  it 
is  returned  when  the  current  again  decreases  to  zero. 

When  the  switch  is  closed  in  the  circuit  in  Fig.  133,  the 
current  rises  in  accordance  with  the  curve  of  Fig.  134,  that 
is,  in  accordance  with  the  equation 

E  *« 

i  =  |(l-6-T).  (36) 

At  any  instant  the  power  input  of  the  circuit  is  equal  to  Ei. 
Of  this  power,  a  part,  Ri2,  is  being  used  up  in  heating  the 
conductor.  The  remainder,  or 

Ei  -  Ri2, 

is  being  stored  up  in  the  magnetic  field.  The  power  going 
into  the  magnetic  field  is  therefore  at  any  instant 

P  =  Ei  -  Ri2 

7/J2  Rt  fi2  Rt  2Rt 

=  ±  (1  -rr)  -  ^-(1  -  2e-T  +  €—) 

J?2          Rt  2Rt 

-f(«~*  -.«"!•).  (37) 

The  total  energy  put  into  the  magnetic  field  is  the  total 
integral  of  this  power,  beginning  at  the  time  when  the  switch 
is  closed;  that  is, 


o 

J?2    /»«         Rt  2Rt 

-C|    (n-«-r)*  (38) 

ttjQ 

which  integrated  gives 

1?2  T         Rt  T.         2Rt  -i°° 

.  ||  w=^~^+m^l-    (39) 

Inserting  the  limits  of  integration,  we  have 


INDUCED    VOLTAGES  271 

or,  since  in  the  steady  state 

E  =  RI,  (41) 

this  may  be  written 

W  =  ^joules.  (42) 

A 

Thus  the  energy  stored  in  an  inductance  of  L  henries 

LP  . 
carrying  a  current  of  /  amperes  is  —=-  joules. 

z 

When  the  electromotive  force  is  removed  from  this  cir- 
cuit, the  current  persists  for  a  certain  length  of  time  as  shown 
in  Fig.  136.  The  energy  stored  in  the  magnetic  field  is 
disappearing,  as  the  current  is  decreasing  to  zero.  The 
voltage  which  forces  the  current  through  the  resistance 
during  this  decrease  of  current  is  simply  the  voltage  of  self- 
inductance.  We  can  compute  the  amount  of  energy  re- 
turned to  the  circuit  during  the  period  in  which  the  current 
is  decreasing  to  zero.  The  energy  returned  to  the  circuit  as 
the  current  dies  to  zero  exactly  equals  the  energy  stored  up 
in  the  magnetic  field  as  the  current  builds  up  from  zero. 
The  equation  of  the  dying  current  is 

Rt 

i  =  Ii~  L.  (43) 

The  rate  at  which  energy  is  being  dissipated  in  heat  is 

mt 
Ri2  =  RI2e~~L,  (44) 

and  therefore  the  total  amount  of  energy  dissipated  in  heat 
as  the  current  dies  out  is 

/»°°  n™  mt 

W  =  /     Ri*dt  =  /     RI2€~~Tdt  (45) 

Jo  Jo 


LI2 


272      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

We  thus  see  that  the  total  amount  of  energy  that  is  stored 
in  the  magnetic  field  during  the  time  that  the  current  is 
being  built  up  is  returned  to  the  circuit  and  appears  in  the 
form  of  heat  during  the  time  that  the  current  is  again  de- 
creasing to  zero. 

We  know  that  in  mechanics,  wherever  there  is  a  mass  M 
moving  at  a  velocity  V,  there  is  represented  a  kinetic  energy 
MF2/2.  Entirely  analogous  to  this,  wherever  in  an  electric 
circuit  there  is  an  inductance  L  carrying  a  current  /,  there 
is  represented  a  stored  energy  LP/2. 

We  have  seen  that  the  magnetic  field  is  really  a  strained 
condition  in  the  ether.  This  strained  condition  represents 
stored  energy.  The  amount  of  this  stored  energy  we  have 
now  learned  to  compute. 

It  thus  takes  energy  to  set  up  a  magnetic  field.  On  the 
other  hand  it  does  not  take  any  expenditure  of  energy  to 
maintain  a  magnetic  field.  When  a  steady  current  is  flow- 
ing through  a  coil  of  wire,  the  energy  input  is  all  used  in  the 
ohmic  loss  of  the  coil  and  appears  as  heat.  No  additional 
energy  is  necessary  to  maintain  the  field.  The  effect  of  the 
inductance,  so  long  as  the  current  is  steady,  is  zero. 

Confusion  sometimes  arises  by  reason  of  the  fact  that 
there  is  always  energy  input  through  a  coil  even  when  the 
current  is  steady.  This  energy,  however,  is  used  up  in 
losses,  and  is  not  used  to  maintain  the  field.  An  analogy 
will  make  this  matter  clear.  A  ten-pound  weight  placed  on 
a  table  represents  potential  energy.  If  the  table  is  three 
feet  from  the  floor,  the  weight  can  do  thirty  foot-pounds  of 
work  in  falling  to  the  floor.  It  thus  represents  a  stored 
energy  of  amount  thirty  foot-pounds.  If  it  is  resting  quietly 
on  the  table,  there  is  no  expenditure  of  energy  to  maintain 
this  potential  energy  at  its  fixed  value.  The  ten-pound 
weight  held  in  the  hand  at  an  equal  distance  above  the  floor 
represents  also  thirty  foot-pounds  of  stored  potential  energy. 
No  expenditure  of  energy  is  necessary  to  maintain  this,  as 
we  have  seen.  Yet  if  one  holds  the  weight  out  at  arm's 


INDUCED    VOLTAGES  273 

length  three  feet  above  the  floor  for  a  considerable  interval, 
his  arm  becomes  decidedly  tired,  and  he  is  conscious  of  ex- 
pending a  considerable  amount  of  muscular  energy.  This 
energy,  however,  is  going  into  muscular  losses  in  his  body, 
and  as  we  have  seen,  is  not  at  all  expended  on  maintaining 
the  potential  energy  in  the  weight. 

Similarly,  a  magnetic  field  always  represents  stored  energy. 
If  this  field  is  due  to  a  permanent  magnet,  it  is  perfectly 
evident  that  no  energy  input  is  required  to  maintain  the 
field.  If  the  field  is  due  to  an  electromagnet,  however,  we 
always  have  an  energy  input  necessary  to  force  the  current 
through  the  resistance  of  the  coil.  This  energy  input  is 
entirely  taken  up  in  ohmic  losses  in  the  coil  and  no  part 
of  it  is  used  in  maintaining  the  stored  energy  in  the 
field. 

If  one  raises  his  arm  holding  the  ten-pound  weight,  and 
lifts  it  to  six  feet  above  the  floor,  he  puts  in  an  additional 
thirty  foot-pounds  of  energy,  making  the  stored  energy  in 
all  sixty  foot-pounds.  In  order  to  do  this  he  must  exert  a 
certain  amount  of  force,  which  means  an  expenditure  of 
muscular  energy.  Besides  the  losses  which  we  have  found 
to  be  present  for  simply  maintaining  the  weight  in  its  fixed 
position,  he  must  expend  a  muscular  energy  of  thirty  foot- 
pounds in  order  to  raise  the  weight  to  its  new  position. 

Similarly,  if  we  increase  the  current  in  a  coil,  we  must  add 
electrical  energy  sufficient  to  increase  the  energy  stored  in 
the  magnetic  field  to  its  new  value.  This  amount  of  energy 
must  be  added  over  and  above  any  incidentally  used  up  in 
PR  losses  during  the  process. 

Prob.  16-8.  Assuming  the  inductance  constant,  what  energy 
is  stored  in  the  field  coil  of  a  generator  where  the  flux  link- 
ing the  coil  is  1.4  X  106  maxwells,  the  number  of  turns  800 
and  the  field-coil  current  11  amperes? 

Prob.  17-8.  The  energy  of  a  coil  carrying  0.3  ampere  is 
2.3  X  10~7  joule.  What  is  the  energy  in  foot-pounds  and 
what  is  the  inductance  in  henries? 


274      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

84.  Energy  in  a  Magnetic  Field.  We  have  seen  that 
when  a  current  of  /  amperes  flows  in  a  coil  of  inductance 
L  henries,  the  total  energy  storage  in  the  field  is 

W  =  ^joules.  (46) 

z 

If  I  is  in  abamperes  and  L  is  in  abhenries,  that  is,  if  we 
are  dealing  with  c.g.s.  electromagnetic  units,  the  energy  will 
be  given  in  ergs;  that  is, 

W=^ferSS.  (47) 

It  should  be  noted  carefully  that  in  deriving  these  for- 
mulas, the  value  of  L  has  been  assumed  constant,  that  is,  we 
assume  that  L  does  not  vary  with  the  amount  of  the  current. 
These  expressions  for  the  energy  in  a  magnetic  field  hence 
hold  strictly  only  for  cases  where  the  reluctance  of  the 
magnetic  circuit  is  constant,  that  is,  when  B  is  proportional 
to  H.  The  formulas  consequently  apply  only  to  coils  in  air, 
the  magnetic  field  about  a  transmission  line  and  so  on.  In 
cases  where  there  is  iron  present  of  varying  permeability, 
these  formulas  must  be  modified.  We  may  sum  this  up  by 
saying  that  the  energy  stored  in  a  magnetic  field  is  as  given 
above  provided  there  is  no  material  of  variable  permeability 
present. 

It  will  be  convenient  to  put  this  formula  into  a  somewhat 
different  form.  Still  dealing  with  constant  permeability, 
we  know  that  the  inductance  is  given  by  the  flux  linkages  per 
abampere  in  the  c.g.s.  system;  that  is, 

L  =^  abhenries.  (48) 

If  A  is  the  cross-sectional  area  of  the  circuit,  assumed  con- 
stant, then  this  may  be  written 

NAB  . 
L  =  — —  abhenries.  (49) 


INDUCED    VOLTAGES  275 

Inserting  this  value  in  expression  (47),  we  obtain 

NAB  I 
W=  — 


Since  from  Ohm's  law  for  the  magnetic  circuit  we  know  that 

D      4irNIfjL  ,     . 

B  =  —  j  —  gausses,  (51) 

which  may  be  written  in  the  form 

7?7 
NI  =  T  —  abampere-turns,  (52) 

we  may  insert  this  expression  in  (50)  and  obtain 

(53) 


However,  the  length  times  the  cross-section  of  the  magnetic 
circuit  is  equal  to  its  volume;   that  is, 

V  =  IA.  (54) 

For  the  energy  per  cubic  centimeter  of  volume  of  the 
magnetic  field,  therefore,  we  finally  obtain  the  expression 

D2 

W  =  5  —  ergs  per  cubic  centimeter.  (55) 

O7T/i 

This  means  that  if  we  establish  a  magnetic  field,  and  the 
flux  density  is  in  gausses,  then  there  is  an  energy  storage 
per  cubic  centimeter  of  this  magnetic  field  which  is  propor- 
tional to  the  square  of  the  flux  density.  Where  the  field  is  in 
air,  n  is  equal  to  unity,  and  we  may  write 

B2 

W  =  5-  ergs  per  cubic  centimeter.  (56) 

O7T 

As  noted,  our  expression  for  the  energy  storage  per  cubic 
centimeter  holds  only  where  the  permeability  is  constant. 
We  shall  see  in  the  next  section  that  the  energy  stored  per 
cubic  centimeter  of  iron  is  always  somewhat  less  than  would 


276      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

be  given  by  this  formula.  For  an  air  gap  in  a  magnetic 
circuit,  however,  the  formula  is  strictly  true.  This  expression 
we  shall  find  convenient  in  dealing  with  electromagnets, 
solenoids  and  so  on. 

Prob.  18-8.  The  air-gap  density  of  a  generator  is  60,000 
lines  to  the  square  inch  and  the  volume  of  the  space  containing 
this  flux  is  13.3  cubic  inches.  What  is  the  amount  of  energy  in 
ergs?  Joules?  Horse-power-hours? 

Prob.  19-8.  If  the  energy  of  the  field  in  Prob.  18-8  were 
dissipated  in  0.001  second,  what  would  be  the  energy  dissipated 

per  second?  Express  the 
power  in  ergs  per  second, 
joules  per  second,  horse 
power. 

Prob.  20-8.  The  cast- 
steel  ring  of  Fig.  139  has 
a  permeability  of  1000 
and  a  flux  density  of 
10,000  gausses.  Make 
no  allowance  for  fringing 
in  the  air  gap  and  com- 
pute the  total  energy  in 
the  air  gap. 

85.  Magnetic  Pull. 
We  have  seen  above 

that  magnetic  flux  lines 
FIG.  139.    A  cast-steel  ring  with  ,  ,  ,       i 

0.5  centimeter  opening.  alwayS  tend  *&<>«** 

in  length.     When  flux 

lines  pass  between  two  iron  surfaces,  they  accordingly  tend  to 
pull  these  surfaces  together.  More  exactly,  the  flux  lines 
passing  completely  around  a  magnetic  circuit  and  tending 
to  shorten  exert  a  force  in  the  direction  tending  to  compress 
the  material  of  the  entire  magnetic  circuit.  If  the  circuit 
is  in  two  parts,  with  an  air  gap  between,  then  the  force  tends 
to  draw  these  two  parts  together.  We  are  all  familiar  with 
this  effect  from  the  tendency  of  electromagnets  or  permanent 
magnets  to  pull  pieces  of  iron  to  themselves.  From  the 


INDUCED    VOLTAGES 


277 


work  of  the  preceding  section  we  are  now  in  a  position  to 
calculate  exactly  the  amount  of  this  force. 

In  Fig.  140  is  shown  a  magnetic  circuit  consisting  of  two 
pieces  of  iron  with  air  gaps  between  of  length  h.  We  will 
assume  that  the  cross-sec- 
tional area  of  each  gap  is 
constant  and  of  value  A 
square  centimeters.  The 
magnetizing  coil,  we  will 
assume,  forces  an  amount 
of  flux  through  the  circuit 
which  gives  a  uniform  flux 


FIG.  140.  The  opening  h  is  increased 
by  an  amount  dh,  by  means  of  the 
force  2F. 


density  of  B  gausses  every- 
where. Let  us  see  what 
the  force  is  in  the  air  gaps 
tending  to  pull  the  two  pieces  of  iron  together. 

The  air  gap,  we  will  assume,  is  short  compared  to  the  cross- 
sectional  area,  so  that  there  is  negligible  fringing.  This 
means  that  the  lines  pass  straight  across  the  air  gap  and  the 
flux  density  is  equal  everywhere  to  B  in  the  air  gap. 

The  energy  storage  in  each  air  gap  will  therefore  be 

W  =  —  V  ergs,  (57) 

which  by  inserting  the  volume  of  the  air  gap  becomes 

W  =  ^-  h  ergs.  (58) 

Let  us  now  assume  that  a  force  is  applied  which  pulls  the 
two  pieces  of  iron  apart  a  certain  additional  small  distance 
which  we  will  call  dh.  This  small  increment  of  air  gap  will 
not  appreciably  affect  the  flux  density.  There  will,  however, 
be  an  additional  volume  in  the  air  gap  and  hence  an  ad- 
ditional energy  storage  of  amount 

B2A 
dw  =  -5 —  dh  ergs.  (59) 

O7T 


278      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Call  the  force  necessary  to  effect  this  separation  2F,  that 
is,  the  force  per  air  gap  F.  Then  the  mechanical  work  done 
in  separating  the  pieces  of  iron  will  be  the  force  times  the 
distance  moved,  or 

Fdh  ergs. 

This  work  done  must  be  equal  to  the  additional  energy  now 
stored  in  the  field;  that  is, 

Fdh  =^dh  ergs.  (60) 


Dividing  by  dh  we  obtain 

P_B*A 

Sw 


dynes.  (61) 


The  magnetic  pull  in  an  air  gap  is  hence  proportional  to 
the  area  and  to  the  square  of  the  flux  density.  If  B  is  in 
gausses  and  A  is  in  square  centimeters,  the  force  will 
be  given  in  dynes,  since  the  dyne  is  the  c.g.s.  unit  of  force. 
Converting  this  expression  to  the  practical  system,  we  ob- 
tain 

F  =  0.014  B2A  pounds,  (62) 

where 

F  is  the  force  in  pounds, 

A  is  the  cross-sectional  area  in  square  inches, 

B  is  the  flux  density  in  kilolines  per  square  inch. 

It  should  be  carefully  noted  that  the  pull  is  proportional 
to  the  square  of  the  flux  density.  For  a  given  amount  of 
flux,  the  greatest  pull  will  therefore  be  obtained  when  this 
flux  is  confined  to  a  small  cross-sectional  area.  An  example 
will  make  this  clear. 

In  Fig.  141  is  shown  a  simple  form  of  lifting  magnet. 
Let  us  assume  that  the  pole  face  PI  has  a  cross-sectional  area 
of  ten  square  inches,  while  P2  has  a  cross-sectional  area  of 
five  square  inches.  Assume  that  the  magnetizing  coil  forces 


INDUCED    VOLTAGES  279 

a  total  amount  of  flux  of  500,000  lines  completely  around 

the  magnetic  circuit.     Computing  such  a  lifting  magnet,  we 

must  be  careful  to  allow  for  an  air  gap  even  when  the  magnet 

comes  closely  in  contact  with  the  sheet  or  other  iron  object 

to  be  lifted,  unless  the  surfaces  are 

very  carefully  brought  together.     If 

there  is  the  slightest  film  of  scale  or 

even  oil  on  the  surfaces  where  they 

touch,  the  resulting  air  gap  may  exert 

a   very    appreciable    effect    on    the 

amount  of  flux  forced  through  the 

circuit.     We  will  assume,  however, 

that  the  coil  supplies  sufficient  mag-    FIG.   141.    A  simple  lift- 

netomotive  force  to  pass  this  total  mg  magnet- 

amount  of  flux  through  the  circuit. 

'  The  flux  density  at  pole  face  PI  will  be  50,000  lines  per 
square  inch,  while  at  pole  face  P2  it  will  be  100,000  lines  per 
square  inch. 

The  total  pull  at  pole  PI  will  therefore  be 

Fl  =  0.014  X  502  X  10  =  350  pounds,  (63) 

while  at  pole  face  P2  the  pull  will  be 

F  =  0.014  X  1002  X  5  =  700  pounds.  (64) 

We  thus  obtain  twice  the  amount  of  pull  at  the  pole  face 
of  small  cross-sectional  area,  even  although  the  same  total 
flux  passes  across  the  air  gap.  We  have  assumed,  of  course, 
that  there  is  no  fringing  of  the  flux.  In  a  practical  example, 
the  above  conclusion  would  be  usually  somewhat  modified 
by  the  effect  of  this  fringing.  For  accurate  results,  however, 
the  effect  of  the  fringing  can  if  desired  be  taken  into  account. 

This  effect,  noted  in  the  example  above,  can  be  illustrated 
practically  in  a  striking  manner.  If  a  magnet  is  arranged  as 
in  Fig.  141  and  a  cord  or  chain  is  applied  to  the  lower  portion 
of  the  circuit  at  .a  fixed  point  M,  it  will  be  found  that  the 
circuit  will  always  open  at  the  face  PI  before  it  will  let  go  at 


280      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

P2.  In  fact,  the  point  of  attachment  would  need  to  be 
moved  toward  PI  until  its  distances  from  the  two  pole  faces 
were  approximately  at  the  ratio  of  two  to  one  before  we  could 
apply  the  maximum  allowable  force  without  separating  the 
two  parts  of  the  magnetic  circuit. 

Advantage  is  taken  of  this  effect  in  lifting  magnets  and 
similar  devices,  by  rounding  pole  faces  where  they  come  in 
contact  with  the  material  to  be  lifted.  If  this  rounding  is 
carried  too  far,  the  reluctance  of  the  magnetic  circle  will, 
of  course,  be  increased  to  the  extent  where  the  resulting 
decrease  in  flux  will  give  a  disadvantage. 

Prob.  21-8.  If  the  core  of  a  transformer  has  a  cross-section 
of  4.2  square  inches  and  a  transverse  gap  of  0.032  inch  is  cut 
so  as  to  be  at  right  angles  to  the  flux  lines,  and  the  flux  density 
of  the  gap  is  9000  gausses,  what  force  in  dynes  acts  to  close  the 
gap? 

Prob.  22-8.  A  simple  "  U  "-shaped  magnet  is  made  of 
a  round  cast-steel  bar  2  inches  in  diameter.  The  mean 
length  of  the  bent  bar  is  22.5  inches.  Across  the  ends  of  the 
magnet  is  placed  a  cast-steel  plate  2  inches  wide,  It  inches 
thick  and  10  inches  long.  Interposed  between  this  plate  and 
each  of  the  ends  of  the  magnet  there  is  a  disk  of  brass  0.021 
inch  thick.  How  many  ampere-turns  will  be  needed  to  sup- 
port 200  pounds  by  means  of  this  magnet  when  the  "  U  " 
portion  of  it  is  held  in  suspension  and  the  weight  is  secured  to 
the  cast-steel  plate? 

86.  Mutual  Induction.  When  a  current  is  varying  in  a 
coil,  there  is  a  voltage  induced  in  the  coil  which  we  call  the 
voltage  of  self-induction.  It  is  equal  to  the  coefficient  of 
self-induction,  L,  times  the  rate  of  change  of  the  current. 
This  coefficient  of  self-induction  we  define  as  the  number  of 
linkages  produced  in  the  coil  by  unit  current. 

In  the  same  manner,  we  may  speak  of  the  mutual  induc- 
tion of  one  coil  upon  another.  The  coefficient  of  mutual 
induction  we  will  call  M.  The  mutual  inductance  of  coil 
1  upon  coil  2  (that  is,  Mi_2)  is  the  number  of  linkages  of  flux 
with  the  turns  of  coil  2  produced  when  there  is  unit  current 


INDUCED    VOLTAGES  281 

passing  in  coil  1.  Similarly,  the  mutual  inductance  of  coil 
2  upon  coil  1  (that  is,  M2-i)  is  equal  to  the  number  of  flux 
linkages  with  coil  1  produced  by  unit  current  flowing  in  coil  2. 
These  two  coefficients  may  be  shown  to  be  equal;  that  is,  the 
mutual  inductance  of  one  coil  upon  a  second  is  equal  to  the 
mutual  inductance  of  the  second  upon  the  first. 

When  the  current  is  varying  in  one  coil,  we  have  shown 
in  Art.  79,  page  250  that  there  is  a  voltage  produced  in  a 
neighboring  coil  through  which  the  flux  links.  This  voltage 
will  be  proportional  to  the  rate  of  change  of  flux  linkages. 
The  voltage  produced  in  the  second  coil  is  thus  equal  to  the 
coefficient  of  mutual  inductance  times  the  rate  of  change  of 
current  in  the  first  coil;  that  is, 

E  =  M%.  (65) 

If  M  is  in  henries  and  the  current  is  in  amperes,  the  voltage 
produced  will  be  measured  in  volts.  Similarly,  if  M  is  in 
c.g.s.  abhenries  and  I  in  abamperes,  E  will  be  in  abvolts. 

When  two  coils  are  situated  close  together,  or  upon  the 
same  iron  core,  so  that  much  of  the  flux  produced  by  one 
will  pass  through  the  other,  we  say  that  the  coils  are  closely 
coupled.  As  a  measure  of  the  closeness  of  coupling,  we  de- 
fine a  coefficient  of  coupling  as  follows.  The  coefficient  of 
coupling  between  two  coils  is  the  coefficient  of  mutual  in- 
ductance between  them,  divided  by  the  mean  self-inductance 
of  the  coils.  We  use  the  geometric  mean  of  the  self  -induc- 
tance in  this  definition.  If  K  is  the  coefficient  of  coupling, 
we  define  K  thus: 


K  = 


When  all  of  the  flux  produced  by  one  coil  will  pass  through 
the  second,  that  is,  when  there  is  no  leakage  flux,  the  coeffi- 
cient of  coupling  is  unity.  Consider  two  identical  coils  with 


282      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

leakage.     Their  coefficients  of  self-inductance  will  then  be 
equal,  or 

Li  =  Lt.  (67) 

Hence 

K-*        or        f-  (68) 

LI  L2 

If  there  is  no  leakage  flux,  however,  the  self-inductance  of 
a  coil  will  be  equal  to  its  mutual  inductance  upon  the  other 
coil,  for  the  same  number  of  linkages  will  be  produced  by  one 
ampere  in  each  case.  In  this  case  the  coefficient  of  coupling 
will  therefore  be  unity.  Suppose  that  the  coils  have  now 
different  numbers  of  turns;  that  is,  assume  that  coil  1  has 
Ni  turns  and  coil  2  has  N*  turns.  Write  the  reluctance  of 
the  magnetic  circuit  connecting  them  as  R.  The  flux  passing 
through  coil  1  is  then 

4,rN1Il 
01  =    ~9T 

so  that  the  self-inductance  of  the  first  will  then  be 

N&rNJ 


(69) 

ZJ\. 

and  of  the  second,  similarly, 
L2  = 


In  the  same  way,  the  coefficient  of  mutual  inductance  be- 
tween the  coils  will  be 


•  (71) 

Comparing  these  three  expressions  we  see  immediately  that 


INDUCED    VOLTAGES 


283 


When  there  is  no  leakage  flux,  then,  the  coefficient  of 
coupling  is  unity. 

In  Fig.  142  are  shown  two  coils  so  arranged  as  to  have 
mutual  inductance.     Flux  lines  are  drawn  for  the  condition 
that  coil  1  is  carrying  a 
current  and  coil  2  is  not. 
It    will    be    noted   that 
some   of    the    flux   lines 
due  to  coil  1  link  coil  2. 
This    shows    that    there 
is  coupling  between  the 
two  coils.      Let  the  in- 


FIG.  142.      Flux  lines  set  up  by 
coil  1  link  coil  2. 


ductances  be  Li  and  L2 
and  M.  Suppose  now 
that  the  two  wires  at  A 

are  connected,  and  that  without  moving  the  coils  we  examine 
the  inductance  of  the  combination  as  measured  by  the  two 
remaining  free  ends.  Assume  that  the  windings  on  the  two 
coils  are  in  the  aiding  direction,  that  is,  so  as  to  force  the 
flux  in  the  same  direction  through  each. 

The  resulting  self-inductance  of  the  combination  will  be 
the  total  linkage  per  ampere  of  the  combination.  A  unit 
current  flowing  in  coil  1  causes  LI  linkages  with  its  own  turns 
and  M  linkages  with  the  turns  of  coil  2.  Similarly,  this  same 
current  flowing  in  coil  2  causes  L2  linkages  with  its  own  turns 
and  M  linkages  with  the  turns  of  coil  1.  The  resultant  self- 
inductance,  that  is,  the  total  flux  linkages  per  ampere  when 
the  coils  are  connected  in  series,  will  be 

L  =  la  +  L2  +  2M.  (73) 

If  the  connection  to  one  of  the  coils  is  now  reversed,  that  is, 
if  the  coils  are  connected  bucking,  the  flux  linkages  due  to 
mutual  inductance  will  be  in  the  opposite  direction  to  those 
produced  by  self-inductance.  In  this  case  the  effect  of 
mutual  inductance  will  therefore  subtract,  and  we  shall 
have  for  the  net  inductance 

L  =  L!  +  L2  -  2M.  (74) 


284      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


This  principle  is  made  use  of  in  an  instrument  called  a 
variometer.  This  is  simply  a  variable  self-inductance.  It 
is  used  in  radio  telegraph  and  telephone  circuits,  for  measure- 
ment work  and  for  purposes  where  a  small  variable  inductance 

is  needed.  The  voltage  regulator 
used  on  alternating-current  circuits 
.  is  a  somewhat  similar  arrangement. 
It  consists  of  two  coils,  as  shown  in 
Fig.  143,  the  outer  one  fixed  and  the 
inner  one  rotatable  about  a  vertical 
shaft  as  shown.  The  two  coils  are 
connected  in  series.  By  rotating 
the  inner  coil,  we  can  vary  the 
mutual  inductance  between  the 

.        .  coils  from  a  maximum,   with  the 

I          I  — 1     coils  aiding,  to  zero,  and  to  a  maxi- 

FIQ.    143.     A  variometer,   mum  with  the  coils  bucking.     The 

self-inductance  of  the  combination 

will  then  be  variable  between  the  limits  given  by  the 
expressions  (73)  and  (74) .  The  range  of  the  device  evidently 
depends  upon  the  ratio  of  M  to  LI  and  L2,  that  is,  upon  the 
coefficient  of  coupling  between  the  two  coils.  In  order  to 
obtain  a  large  range  for  the  instrument,  this  coefficient  of 
coupling  is  kept  as  large  as  possible.  The  coils  are  for  this 
purpose  placed  with  little  clearance  between  them  when  in 
the  same  plane,  in  order  that  there  may  be  little  leakage  of 
flux.  Turning  the  inner  coil  180°  varies  the  inductance  of 
the  combination  from  its  minimum  to  its  maximum  value. 
When  the  coils  are  at  right  angles  as  shown  in  the  figure,  there 
is  no  mutual  induction  between  them,  and  the  self -inductance 
of  the  combination  is  simply  the  sum  of  the  self-inductances 
of  the  two  coils  alone. 

When  studying  coefficients  of  self-induction,  we  noted  that 
the  self-inductance  of  a  coil  was  a  constant  only  when  the 
permeability  of  the  magnetic  circuit  in  question  was  also  con- 
stant, that  is,  when  there  was  no  iron  present.  With  iron 


INDUCED    VOLTAGES  285 

present,  the  self-inductance  depends  upon  the  value  of  the 
current  at  which  it  is  measured.  Similarly  in  the  case 
of  mutual  induction,  the  coefficient  of  mutual  induction 
between  two  coils  is  a  fixed  value  only  when  the  permeability 
of  the  magnetic  circuit  connecting  them  is  a  fixed  value. 
When  iron  is  present,  the  coefficient  of  mutual  induction  is 
dependent  upon  the  amount  of  current  in  the  coil  at  the  time 
that  the  coefficient  is  measured,  and  the  coefficient  no  longer 
has  a  definite  meaning. 

Prob.  23-8.  A  concentrated  coil  of  200  turns  of  small  wire 
(small  to  cut  down  leakage)  has  an  inductance  of  0.010  henry. 
The  number  of  turns  is  increased  to  400.  What  will  be  the 
approximate  change  in  inductance? 

Prob.  24-8.  Two  coils  have  inductances  of  3.20  henries  and 
2.10  henries  respectively  and  the  mutual  inductance  is  2.00 
henries.  What  is  their  coefficient  of  coupling?  Sketch  the 
coils  in  position  for  maximum  and  for  minimum  mutual  in- 
ductance and  show  the  direction  of  current  in  each  for  the  two 
conditions. 

Prob.  25-8.  Two  inductances  are  connected  in  series  in 
such  a  way  that  they  have  a  total  inductance  of  0.20  henry. 
The  coils  have  self-inductances  of  0.05  and  0.09  henry  re- 
spectively. What  is  the  mutual  inductance  and  what  would 
be  the  inductance  of  the  two  coils  in  series  if  the  leads  of  one 
were  reversed  and  their  position  in  respect  to  each  other  un- 
altered? 

Prob.  26-8.  It  is  often  necessary  to  construct  an  inductance 
in  such  a  way  that  its  value  may  be  calculated.  (For  details 
see  "  Construction  and  Calculations  of  Standards  of  Induc- 
tance," Bulletin,  U.  S.  Bureau  of  Standards,  Vol.  2,  pp.  87-143, 
1906.)  We  know  that  for  a  maximum  inductance  a  given 
amount  of  wire  should  be  wound  in  a  channel  of  square  cross- 
section  and  the  mean  radius  of  the  coil  should  be  1.85  times  the 
length  of  a  side  of  the  channel.  The  self-inductance  of  such  a 
coil  is  expressed  by 

L  =  19.347  an*  10~9  henries, 
where 

a  =  the  mean  radius  of  the  coil  in  centimeters, 
n  =  the  number  of  turns. 


286      PRINCIPLES  OF  ELECTRICAL  ENGINEERING' 

Design  a  coil  for  use  as  a  standard,  of  inductance  10  millihenries. 
This  inductance  coil  is  to  be  used  for  low  frequencies,  con- 
sequently it  may  be  made  of  No.  18  d.c.c.  (double  cotton  cov- 
ered) wire,  diameter  over  cotton  0.048  inch. 


SUMMARY    OF   CHAPTER  VIII 

LENZ'S  LAW  STATES  THAT:  — 

(a)  Whenever  there  is  a  change  in  the  amount  of  magnetic 
flux  linking  an  electric  circuit,  a  voltage  is  set  up  tending  to 
produce  a  current  in  such  a  direction  as  to  oppose  this  change 
in  flux  ; 

(6)  The  voltage  thus  set  up  is  directly  proportional  to  the 
rate  of  change  of  flux  linkages. 

As  an  equation  this  is  expressed  as  follows  :  — 

E  =  N  ^  10-8  volts. 
at 

IF  THE  FLUX  VARIES  HARMONICALLY  with  the 
time,  then 

B    =  Bma*  Sin  27T/J, 

where 

/  =  the  frequency  in  cycles  per  second, 
t  =  the  time  in  seconds. 

The  voltage  will  then  be 

E  =  27rNABmax  f  cos  2?r  ft. 

THE  COEFFICIENT  OF  SELF-INDUCTION  is  the  num- 
ber of  linkages  set  up  by  a  coil  with  its  own  turns  when  the  coil 
is  carrying  unit  current. 

THE  VOLTAGE  OF  SELF-INDUCTION  IS 


where 

L  =  the  coefficient  of  self-induction. 

THE  GENERAL  LAW  CONCERNING  THE  DISTRI- 
BUTION OF  VOLTAGES  in  an  electric  circuit  containing 
resistance  and  inductance  is  as  follows. 

287 


288      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

At  any  instant  the  sum  of  the  resistance  drops  and  induced 
voltages  equals  the  impressed  voltage  at  that  instant.  The 
general  form  of  the  equation  is 

iR+LjJ-B.     .-  .;- 

WHEN  THE  SOURCE  OF  THE  ELECTROMOTIVE 
FORCE  IS  SHORT  CIRCUITED,  this  equation  becomes 


IF  THE  IMPRESSED  VOLTAGE  IS  SUDDENLY  DOU- 
BLED or  if  it  is  reduced  one-half,  the  equation  becomes  re- 
spectively 


and 


iR  +L       =  2E 
at 


dt       2 


THE  CURRENT  IN  A  CIRCUIT  CONTAINING  RESIST- 
ANCE R  AND  INDUCTANCE  L  GROWS  according  to  the 
equation 

E  Rt 

- 


THE   CURRENT   DECREASES   IN  THE    SAME    CIRCUIT 
when  the  e.m.f.  is  removed  according  to  the  equation 

Rt 

1    =  l€      L  • 

THE  TIME  CONSTANT  of  such  a  circuit  is  the  time  in 
which  the  current  will  decrease  to  1/e  of  the  original  value  when 
the  e.m.f.  is  removed;  or  it  is  the  time  in  which  the  current 
would  reach  zero  if  it  continued  to  decrease  at  its  initial  rate. 

MAGNETIC  ENERGY  is  analogous  to  the  energy  of  moving 
bodies. 

MV2 

Kinetic  energy  =  -  foot-pounds. 
2 

LI2 
Magnetic  energy  =  —  ergs. 


INDUCED    VOLTAGES  289 

THE  ENERGY  STORED  IN  A  MAGNETIC  FIELD  (con- 
stant permeability)  equals 

B2 

W  = ergs  per  cubic  centimeter. 

OTTfJL 

THE  MAGNETIC  PULL  can  be  found  from  the  equation 
F  ==^dynes, 

where 

B  =  the  flux  density  in  the  air  gap  in  gausses, 
A  =  the  area  of  the  air  gap  in  square  centimeters. 

THE  COEFFICIENT  OF  MUTUAL  INDUCTION  is  the 
number^of  changes  in  flux  linkages  made  in  Coil  2  when  Coil 
1  changes  its  current  value  one  ampere. 

4x  NiN2 

JVL    =  79; * 


COILS  ARE  SAID  TO  BE  CLOSELY  COUPLED  when 
nearly  all  the  flux  which  threads  one  coil  also  threads  the 
other.  The  measure  of  this  coupling  is  expressed  by  the 
equation 

M 

K  =     , » 

VLXL2 

where 

K  =  the  coefficient  of  coupling, 

LI  =  the  self-inductance  of  Coil  1, 

L2  =  the  self-inductance  of  Coil  2, 

M  =  the  mutual  inductance  of  the  coils. 

FOR  TWO  COILS  CONNECTED  IN  SERIES  with  their 
magnetomotive  forces  in  the  same  direction,  the  self-induct- 
ance of  the  combination  may  be  expressed  by  the  equation 

L  =  Li  +  L2  +  2M. 

If  the  magnetomotive  forces  of  the  coils  are  opposed,  the 
equation  for  self-inductance  of  the  combination  becomes 

li  =  L!  +  L2  -  2M. 


PROBLEMS   ON   CHAPTER   VIII 

Prob.  27-8.  A  series  circuit  consisting  of  a  15- volt  battery 
with  an  internal  resistance  of  1.6  ohms,  a  resistance  coil  (non- 
inductively  wound)  of  10  ohms  and  an  inductance  of  0.10 

77!  1   P\ 

henry  is   carrying   its   normal   current  —  =  -   -  =  1.293   am- 

H       11. o 

peres.  Half  of  the  cells  of  the  battery  are  suddenly  short 
circuited.  Show  on  a  plot  the  original  constant  current  and 
the  transient  current  caused  by  the  short  circuiting  of  the  cells. 

Prob.  28-8.  Four  pancake  inductance  coils  are  to  be  con- 
nected in  series  and  placed  in  a  square  box  so  as  to  make  a 
standard  of  inductance.  How  would  you  place  these  coils 
in  the  box  so  that  their  mutual  inductance  would  be  a  mini- 
mum? Could  you  reduce  the  effect  of  mutual  inductance  to 
zero? 

Prob.  29-8.  Loading  coils  are  used  in  telephone  lines  to 
enable  the  speech  to  be  transmitted  over  long  distances.  (See 
"  Telephonic  Transmission,"  by  J.  G.  Hill.)  These  coils  are 
iron-cored  and  toroidal  in  form.  The  ratio  of  R  to  L  is  ap- 
proximately as  follows : 

D 

For  the  "  loading  "  of  aerial  lines  —  =  25. 

u 

r> 

For  the  "  loading  "  of  underground  lines  —  =  50. 

h 

This  ratio  can  be  held  nearly  constant  and  is  determined  by 
measuring  the  inductance  when  a  current  of  0.8  milliampere 
flows  in  the  coil.  For  aerial  loading,  the  inductance  "  L  " 
is  usually  about  260  millihenries  to  the  coil  and  the  spacing 
is  such  that  there  is  one  coil  about  every  8  miles.  Taking  a 
soft-steel  core  of  circular  section  (see  Fig.  112)  where  rz  —  rx  = 
0.75  inch  and  r  =  1.5  inches,  design  a  winding  on  the  coil 
for  an  inductance  of  260  millihenries  when  the  current  I\  in 
the  winding  is  0.8  milliampere. 

Prob.  30-8.  A  General  Radio  Company  variometer  (Type 
190-650-59)  has  the  following  constants: 

290 


INDUCED    VOLTAGES  291 

Stator  turns  1174, 
Rotor  turns  1174, 
Total  inductance  with  the  two  coils  in  series  aiding  (fields 

in  the  same  direction  and  coils  in  the  same  plane)  626.5 

millihenries, 
Total  inductance  of  the  two  coils  in  series  opposing  (fields 

in  opposite  directions  and  the  coils  in  the  same  plane) 

106.5  millihenries. 

What  is  the  mutual  inductance?     The  energy  stored  when 
»  =  0.2  ampere  with  coils  aiding?     With  coils  opposing? 

Prob.  31-8.  An  artificial  transmission  line  in  the  Elec- 
trical Engineering  Research  Laboratory,  Massachusetts  In- 
stitute of  Technology,  is  made  up  of  twenty-six  sections  in 
series.  Each  section  has  an  inductance  of  0.0567  henry  and 
a  resistance  of  3.359  ohms.  If  a  voltage  of  110  is  applied  to 
the  line  (the  resistance  and  the  inductance  of  the  generator  being 
negligible),  one  lead  connected  to  each  end  of  the  line,  what  will 
happen  if,  after  the  current  has  reached  a  constant  value,  50  ohms 
is  suddenly  inserted  in  series  with  the  line  and  potential  source? 
Plot  the  current  from  its  original  value  to  the  value  it  has  with 
the  additional  resistance 
in  series.  What  is  the 
total  magnetic  energy  of 
the  line  before  inserting 
the  resistance?  After? 


Prob.  32-8.    A  variable 
inductor        ( variometer ) 
such    as    the    type     107, 
General  Radio  Company,        FIG.  144.     A  toroidal  repeating  coil 
has  with  its  coils  in  series  used  in  telephone  circuits, 

aiding   and    opposing   in-    • 

ductances  of  0.6  and  0.12  millihenry  respectively.     What  is  the 
coefficient  of  coupling  "&"?    L  is  same  for  both  coils. 

Prob.  33-8.  A  toroidal  repeating  coil  (see  Fig.  144)  used 
in  ordinary  telephone-cord  circuits  of  central  energy  systems 
(Fender,  "  American  Handbook,"  p.  1542,  Fig.  29)  was 
measured  for  inductance.  One  of  the  four  windings  of  the 
coil  had  an  inductance  of  240  millihenries  when  measured 
by  a  standard  inductance  bridge.  The  same  winding  when 
measured  with  a  current  of  one  ampere  flowing  had  an  in- 
ductance of  0.523  millihenry.  Explain  the  discrepancy. 


292      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  34-8.  Telephone  loading  coils  are  placed  one  on  top 
of  another  and  concentrically  in  iron  "  pots."  In  this  way 
they  are  placed  where  they  are  to  be  used  on  the  line.  When 
the  coils  are  put  into  place  in  the  pots,  they  are  separated  by 
thin  sheet-iron  disks.  What  is  the  reason  for  taking  this 
precaution? 

Prob.  35-8.  In  using  a  ballistic  galvanometer  as  a  flux 
meter  for  measuring  fluxes  of  large  values,  why,  if  you  wanted 
to  reduce  the  galvanometer  deflection,  would  you  put  a  re- 
sistance in  series  with  the  galvanometer  and  not  in  parallel? 

Prob.  36-8.  The  energy  stored  in  an  inductive  circuit  is 
equal  to  \  Li2,  and  the  rate  of  energy  dissipation,  i2r.  It  is 
sometimes  desired  to  keep  the  ratio  of  i  Li2  to  i2r  large.  In  the 
case  of  the  toroidal  repeating  coil,  Fig.  144,  the  total  inductance 
of  all  the  coils  in  series  aiding  is  960  millihenries,  the  total  re- 
sistance 92  ohms.  What  is  the  energy  storage  if  the  rate  of 
energy  dissipation  is  5  X  10~5  watts?  (Such  a  value  of  en- 
ergy storage  though  small  is  characteristic  of  telephone  cir- 
cuits.) 

Prob.  37-8.  The  battery  current  is  sent  to  the  telephone 
instruments  of  the  subscribers  by  means  of  the  repeating  coil 
connections  as  shown  in  Fig.  144.  What  effect  has  this  flow 
of  current  on  the  inductance  of  the  repeating  coil  if  the  lines 
to  the  subscribers  are  both  of  the  same  resistance? 

Prob.  38-8.  Christie  gives  the  following  design  of  a  250- 
kilowatt,  250-volt,  8-pole,  400-revolutions-per-minute  direct- 
current  generator  ("  Electrical  Engineering,"  Christie):  flux 
per  pole  7.08  X  106  lines,  number  of  turns  in  field  coil  960, 
field  current  7.5  amperes,  resistance  of  each  field  coil  3.35  ohms. 
Assuming  that  all  the  flux  links  all  the  turns  and  that  the  in- 
ductance remains  constant,  how  long  after  a  voltage  has  been 
applied  to  the  coil  would  an  ammeter  read  63%  of  the  final 

Tjl 

constant  current  value   — ?      With  the  above  assumptions,  if 
H 

the  field  circuit  were  opened  with  the  eight  pole  windings  in 
series,  what  total  energy  would  be  dissipated  in  the  resistance 
of  the  winding  and  the  spark  at  the  switch  points? 

Prob.  39-8.  In  a  Cutler  circuit -breaker  (Cutler-Hammer 
Company  circuit- breaker,  Type  W),  Fig.  145,  rated  at  100 
amperes  and  200  volts,  the  weight  of  A- A  is  0.4  pound. 
Find  the  flux  density  in  the  gap  C'-C  necessary  to  lift  the 
armature  A- A.  Assume  that  half  the  weight  of  A- A  is  sup- 


INDUCED    VOLTAGES 


293 


ported  by  the  hinge  and  consider  that  the  air  gap  is  of  the  same 
area  as  the  shoe  C"  and  that  the  magnetic  circuit  is  of  soft  iron. 

Prob.  40-8.     If,  in  Prob.  39-8,  52%  of  the  flux  due  to  a  coil 
of  9  turns  wound  on  the  core  Bi-Bi  fails  to  pass  through  the 
gap   C'-C,   what    must    be 
the  value    of    the    current 
in  the  coil  to  lift  the  latch 
A-A,     Fig.     145?        (The 
reluctance  of  the  iron  path 
is     very    small     compared 
with  that  of  the  air  gap.) 

Prob.  41-8.  A  10-horse- 
power,  220-volt  motor  has 
a  starting  box  whose  con- 


A  magnetic  circuit  - 
breaker. 


FIG.  145. 

tact  arm  is  equipped  with 
a  no-voltage  release.     The 

coil  of  the  release  is  in  series  with  the  motor  field,  and  is 
wound  on  the  portion  B-B  of  the  soft  iron  magnetic  path  shown 
in  Fig.  146.  If  the  spring  attached  to  the  starting  arm  exerts 

a  pull  of  0.4  pound  on  the 
center  of  the  armature  A 
perpendicular  to  axis  of 
B-B,  at  what  value  of 
voltage  will  the  device 
operate?  The  normal  field 
current  is  1.6  amperes  and 
the  release  coil  has  100 
turns  of  wire. 

Prob.   42-8.       A     series 


lh" >| 


FIG.  146.  A  spring  attached  to  the 
center  of  A  pulls  it  away  from  the 
magnet  when  the  flux  density  drops 
below  a  certain  value. 


contactor  (General  Electric 
Review,  Vol.  15,  p.  261, 
1912)  is  made  entirely  of 
soft  iron.  The  core  is  round 
and  is  wound  with  300  turns 

of  No.  18  d.c.c.  copper  wire.     With  a  current  of  3  amperes, 
what  pull  will  be  produced?     See  Fig.  147. 

Prob.  43-8.  The  total  flux  per  pole  from  the  permanent 
magnets  alone  of  a  standard  bipolar  telephone  receiver  is  311 
maxwells.  The  area  of  each  pole  face  is  0.199  by  1.14  centi- 
meters. With  this  flux,  what  is  the  pull  on  the  receiver  dia- 
phragm, assuming  that  all  this  flux  passes  from  the  pole  faces 
into  the  diaphragm  perpendicularly?  If  a  voice  current  in- 


294       PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

creases  the  total  flux  by  5  maxwells,  what  is  the  change  in  pull 
on  the  diaphragm?  If  the  effect  of  the  permanent  magnets 
were  removed,  and  with  the  same  pole  faces  and  diaphragm, 
what  would  be  the  pull  exerted  on  the  diaphragm  by  5  max- 
wells? 


FIG.   147.    A  series  contactor. 

Prob.  44-8.  Design  the  winding  for  the  magnetic  chuck 
of  Fig.  148  so  that  it  will  hold  to  its  face  in  intimate  contact 
with  a  total  force  of  1200  pounds  a  cast-iron  billet  or  bar  of 
the  dimensions  shown.  The  material  used  throughout  the 
construction  of  the  chuck  is  cast  steel.  Allowable  watts  per 


•J&.J 


£ 


Cast  iron  billet  placed 
as  shown  by  dotted  lines 


t 


Winding/ 

Section  X-X 

FIG.  148.    A  magnetic  chuck. 

square  inch  of  coil  (curved)  radiating  surface  0.5.  The  impressed 
voltage  is  230.  Neglect  leakage  in  the  air  gap.  Select  the  size  of 
wire,  determine  the  total  length  by  rinding  the  mean  length  of 
one  turn  and  multiplying  by  the  number  of  turns  (see  "Ameri- 
can Machinist,"  1915,  article  by  Fish). 


INDUCED    VOLTAGES 


295 


FIG.  149.     An  annular  magnetic 
chuck. 


Prob.  45-8.  The  annular  chuck  of  Fig.  149  is  to  be  oper- 
ated on  110  volts  to  hold  to  its  surface  with  an  average  force  of  50 
pounds  to  the  square  inch  a  disk  of  cast  steel  of  the  same  dia- 
meter as  the  chuck  and  0.75  inch  thick.  The  factor  of  safety 
shall  be  one  and  one-half.  Allowable  watts  radiation  per  square 
inch  of  radiating  surface  (surface  not  in  contact  with  the  disk) 
0.5  watt.  Design  the  winding.  (References:  "  American 
Machinist,"  1915,  article 
by  Clewell;  "  Electrical 
World,"  1919,  article  by 
Kenyon.) 

Prob.  46-8.  An  induct- 
ance coil  having  1  ohm 
resistance  and  2  henries 
inductance  is  connected 
in  series  with  a  4-ohm 
non-inductive  resistance. 

If  at  zero  time  a  volt- 
age of  50  volts  is  im- 
pressed upon  this  circuit 
and  0.1  second  later  this  voltage  is  suddenly  increased  to  100 
volts,  how  long  will  it  take  for  the  current  to  reach  95  per- 
cent of  its  final  value? 

Prob.  47-8.  A  telegraph  relay  having  a  resistance  of  10 
ohms  and  an  inductance  of  1  henry  is  operated  by  a  24-volt 
storage  battery.  The  iron  is  worked  at  such  a  low  flux  density 
that  the  permeability  may  be  considered  constant  with  only 
slight  error.  To  the  armature  of  this  relay  is  attached  a  con- 
tact which  connects  a  non-inductive  resistance  of  1  ohm  across 
the  terminals  of  the  relay  when  the  armature  of  the  relay  picks 
up,  and  disconnects  this  shunt  resistance  when  the  armature 
drops.  The  relay  is  permanently  connected  to  the  battery 
through  a  resistance  of  8  ohms.  The  armature  is  so  light  that 
inertia  effects  may  be  neglected.  The  armature  picks  up  when 
the  current  reaches  a  value  of  1  ampere  and  drops  when  the 
current  falls  to  0.5  ampere.  Will  the  relay  vibrate?  If  so, 
what  time  is  required  for  the  relay  to  go  through  a  complete 
cycle? 

Prob.  48-8.  A  circuit  containing  inductance  and  resistance 
has  10  volts  impressed  upon  it  and  the  resulting  current  read 
by  an  ammeter  is  0.10  ampere.  The  growth  of  this  current 
in  the  circuit  is  shown  in  Fig,  150,  which  is  a.n  oscillogram  taken 


296      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

at  M.I.T.     The  time  is  given  by  the  60-cycle  wave  at  the  bot- 
tom.    What  is  the  inductance  of  the  circuit? 


FIG.  150.  An  osefllogram  of  the  growth  of  a  current  in  a  circuit  con- 
taining inductance  and  resistance.  Taken  by  Prof.  F.  S.  Dellenbaugh> 
Research  Division,  Electrical  Engineering  Dept.,  M.I  .T. 


FIG.  151.    An  oscillogram  of  the  decay  of  a  current  in  the  circuit  of 
Fig.  150.     Taken  by  Prof.  F.  S.  Dellenbaugh,  M.I.T. 

Prob.  49-8.  Fig.  151  is  an  oscillogram  taken  at  M.I.T.  of 
the  decay  of  current  in  the  same  circuit  as  Fig.  150.  The  coil 
was  short  circuited  and  the  power  taken  off,  the  constants  of 


INDUCED    VOLTAGES 


297 


the  circuit  and  current  scale  during  the  decay  of  current  being 
the  same  as  during  the  growth  investigated  in  the  last  problem. 
Show  by  plotting  the  synthesized  curve  from  the  equation  that 
this  curve  checks  equation  23,  page  263. 

Prob.  50-8.  In  Fig.  151a  the  inductance  coil  (L)  has  an 
inductance  of  50  millihenries  and  a  resistance  of  2  ohms;  the  re- 
sistance (R)  is  five  ohms.  The  battery  has  an  e.m.f.  of  20  volts 
and  an  internal  resistance  of  0.25  ohm. 

(a)  If  the  key  (K)  is  suddenly  closed,  determine  the  values  of 
the  currents  IR,  IE  and  IL  at  any  time  (t).  (6)  What  is  the 
current  in  each  at  0.0005  second?  at  0.005  and  at  0.05  second? 
After  conditions  have  become  steady,  the  key  is  opened,  (c) 
If  it  were  possible  to  open  this  switch  with  absolutely  no  spark, 
what  would  be  each  current  the  instant 
after  opening  and  at  the  end  of  0.002 
second?  (d)  What  would  be  the  volt- 
age across  L  immediately  after  open- 
ing K? 

Prob.  51-8.  A  transmission  line 
is*  made  up  of  two  parallel  conductors 
of  750,000  circular  mils  cross-section, 
separated  by  a  distance  of  10  feet.  The 
line  is  170  miles  long.  Compute  the 

magnetic  energy  stored  in  the  space     FIG.  151a.    A  circuit  con- 
surrounding    the    line,    when     it     is         taing  resistance  and  in- 
carrying   a   current  of  700  amperes.         ductance. 
If    the   circuit    is    interrupted    when 
carrying  this  current,  what  must  become  of  this  energy? 


CHAPTER  IX 
THE  MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL 

Iron  and  steel  are  practically  the  only  materials  available 
for  the  construction  of  magnetic  circuits.  No  other  material 
approaches  these  in  permeability.  We  shall  consider  in 
this  chapter  the  magnetic  properties  of  iron  and  steel. 

87.  Magnetic  Retentivity  and  Hysteresis.  We  have 
already  seen  that  the  permeability  of  iron  depends  upon  the 
extent  to  which  it  is  magnetized.  The  permeability  of 
ordinary  iron  varies  from  one  hundred  or  so  for  very  low 
values  of  the  magnetizing  force  to  three  or  four  thousand 
for  moderate  degrees  of  magnetization  and  then  decreases, 
so  that  at  very  high  flux  density  the  permeability  becomes 
very  low  indeed.  This  is  strikingly  shown  in  Fig.  152, 
in  which  are  curves  drawn  from  data  taken  on  sheet  steel 
by  Dr.  Miles  Walker,  Manchester,  England.  Note  that  for 
a  flux  density  of  25,000  gausses,  a  magnetizing  force  of  2600 
ampere-turns  per  centimeter  or  3260  gilberts  per  centimeter 
is  required.  The  permeability  of  the  steel  at  this  density  is 
only  25,000/3260  or  7.7. 

If  the  magnetizing  force  is  sufficiently  increased,  the  per- 
meability will  approach  unity;  that  is,  the  effect  of  the 
presence  of  iron  is  almost  entirely  lost,  and  the  additional 
flux  which  is  forced  through  by  very  high  magnetizing  forces 
encounters  a  reluctance  almost  as  great  as  would  be  present 
were  the  magnetic  circuit  of  air  alone.  We  say  that  the  iron 
is  then  saturated. 

Not  only  is  the  value  of  the  permeability  dependent  upon 
the  flux  density  but  there  is  a  certain  further  effect  present. 
The  flux  density  which  will  be  set  up  by  a  given  magnetizing 
force  depends  upon  how  that  force  is  applied;  that  is.  whether 

298 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     299 


^^ 

^^ 

-— 

— 

;==- 

' 

^** 

^^ 

^^ 

6C 

o  x 

/ 

2C 

0     4C 

y 

X    8C 

K)    10 

00  12 

00  14 

W    16 

00  18 

M  20 

yo  22 

00   24 

00  26 

X) 

/ 

7- 

,^—  •— 

—  ' 

^^ 

^^ 

,•*•  — 

,  - 

/ 

^ 

^ 

-^-"" 

6 

^X 

•^ 

20       4 

y 

X 

8 

D      1C 

0     U 

0     1^ 

0      1€ 

0     11 

0      2( 

K)     2i 

0     2' 

0      2( 

0 

i 

7 

+  — 

•" 

^  — 

^* 

^  • 

,  • 

„.     •" 

—    •• 

/ 

x^ 

> 

X 

/ 

I 

I/ 

f 

/ 

/ 

0       2      4       6       8      10     12     14     16     18     20     22     24     26 

Ampere-turns  per  Centimeter 

FIG.  152.     Magnetization  curve  of  sheet  steel  at  unusually  high  flux 
densities.     Courtesy  of  Dr.  Miles  Walker,  Manchester,  England. 


300      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

it  is  increased  to  its  final  value  or  has  previously  been  raised 
higher  and  then  decreased.  The  iron  tends  to  retain  mag- 
netism when  it  is  once  set  up  in  it.  In  fact,  if  the  magnet- 
izing force  is  applied  to  a  specimen  of  iron  and  the  force  is 
then  removed,  the  iron  will  be  found  to  retain  a  certain 
fraction  of  its  magnetism.  This  is  called  the  residual  mag- 
netism of  the  specimen.  We  are  familiar  with  this  effect 
in  the  preparation  of  permanent  magnets.  It  is  much  more 
marked  with  hard  steel  than  with  soft  iron.  A  piece  of  soft 
iron,  strongly  magnetized  by  a  coil  and  then  removed  from 
the  vicinity  of  the  magnetizing  force,  will  be  found  to  retain 
but  little  permanent  magnetism.  A  piece  of  hardened  steel, 
on  the  other  hand,  may  retain  as  much  as  fifty  to  eighty  per- 
cent of  the  maximum  flux  set  up  in  it.  The  residual  magnet- 
ism of  a  sample  of  iron  or  steel  is  the  flux  density  which  it 
retains  after  being  magnetized  and  having  the  magnetizing 
force  slowly  removed.  In  order  to  remove  this  residual 
magnetism  it  is  necessary  to  apply  a  magnetizing  force  in 
the  opposite  direction.  The  magnetizing  necessary  in  order 
to  bring  the  specimen  back  to  a  state  where  it  is  entirely 
demagnetized  is  called  the  coercive  force.  See  Fig.  153. 

The  residual  magnetism  (flux  density)  of  a  specimen  after 
being  saturated  is  called  the  retentivity  of  the  iron.  The 
residual  magnetism  of  a  given  kind  of  iron  or  steel  will 
depend  upon  the  maximum  flux  density  used  in  the  test  and 
also  upon  the  shape  of  the  specimen.  A  ring-shaped  speci- 
men uniformly  magnetized  will  retain  much  more  residual 
magnetism  than  a  straight  bar  of  the  same  material.  The 
retentivity  of  a  material  should  be  measured  on  a  ring 
specimen.  The  reason  that  a  bar  will  retain  less  mag- 
netism is  on  account  of  the  demagnetizing  effect  of  its  poles. 
This  matter  will  be  considered  again  later. 

We  have  seen  that  when  a  magnetic  circuit  is  carrying  a 
flux  of  constant  magnitude,  there  is  no  heating  of  the  iron. 
However,  it  is  found  that  if  the  flux  is  varying  in  amount  or 
in  direction,  the  iron  becomes  heated,  thus  showing  that 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL      301 

there  is  a  loss  occurring.  This  loss  is  due  to  the  fact  that 
the  flux  density  does  not  strictly  follow  the  magnetizing 
force  but  lags  behind  it  in  value.  A  certain  amount  of 
energy  which  appears  as  heat  is  thus  used  up  in  the  specimen. 
This  effect  is  known  as  hysteresis  and  the  loss  which  appears 
as  heat  is  known  as  hysteresis  loss  when  it  is  due  to  this 
cause. 

The  reason  for  the  appearance  of  hysteresis  loss  can  be 
made  clear  by  means  of  an  analogy.  Suppose  that  an 
automobile  is  coasting  down  a  hill  with  its  gears  enmeshed 
and  its  clutch  engaged,  so  that  the  engine  is  turning  over. 
Suppose  that  the  throttle  is  entirely  closed.  In  an  actual 
machine  it  is,  of  course,  not  possible  to  close  the  inlet  entirely 
without  special  adjustment.  If  there  were  no  mechanical 
friction  in  the  engine  and  the  transmission,  the  rotation  of 
the  engine  would  then  offer  no  resistance  to  the  progress  of  the 
car.  Air  would  be  compressed  in  the  cylinders  during  the 
compression  stroke  and  would  expand  during  the  stroke 
which  would  normally  be  a  working  stroke.  If  the  valves 
do  not  leak  and  there  is  no  other  leakage  passage  open,  the 
amount  of  work  done  upon  the  gas  in  compressing  it  is  ex- 
actly returned  when  the  gas  expands.  However,  if  one  of 
the  valves  leaks  a  little,  the  pressure  on  the  expansion  stroke 
is  less  than  at  the  corresponding  point  on  the  compression 
stroke.  As  a  result,  not  all  of  the  energy  of  compression  is 
returned  on  the  expansion  stroke,  and  the  engine  takes  power 
to  drive  it  and  hence  slows  down  the  car. 

The  same  thing  is  true  in  a  magnetic  circuit.  If  the  flux 
density  exactly  follows  the  magnetizing  force  at  every  point, 
all  of  the  work  stored  in  the  magnetic  field  as  the  flux  in- 
creases will  be  returned  to  the  circuit  when  the  flux  decreases 
and  no  loss  will  occur.  Due  to  the  retentiveness  of  the  iron, 
however,  the  flux  will  not  exactly  follow  the  magnetizing 
force  applied  and  a  loss  will  occur  whenever  the  flux  density 
varies.  If  the  flux  density  varies  through  a  complete  cycle 
of  changes  and  returns  to  its  original  value,  a  certain  net 


302      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


amount  of  loss  will  be  converted  into  heat,  called  the  hys- 
teresis loss  per  cycle. 

In  a  transformer  we  have  seen  that  the  flux  varies  periodic- 
ally between  certain  maximum  limits.  In  the  iron  of  the 
transformer  core  there  is  hence  a  core  loss  due  to  hysteresis. 
There  is  also  an  additional  core  loss  due  to  eddy  currents 
which  we  will  study  in  the  next  chapter.  The  total  hysteresis 
power  loss  will  be  equal  to  the  hysteresis  loss  per  cycle  times 
the  number  of  cycles  per  second.  If  the  constants  for  hys- 
teresis loss  apply  to  one  cubic  centimeter  of  material,  we 
must  then  multiply  the  result  by  the  volume  of  the  core  in 
cubic  centimeters. 

An  electric  generator  or  motor  consists  of  an  iron  armature 
which  rotates  in  a  magnetic  field.  The  flux  in  such  an 

armature  is  thus  varying 
continuously  in  direction 
relative  to  the  iron.  It 
has  been  found  experimen- 
tally that  the  same  loss 
occurs  due  to  hysteresis, 
when  the  direction  of  a  flux 
changes  continuously  in  this 
manner,  as  will  occur  when 
FIG.  153.  A  hysteresis  loop.  the  direction  of  the  flux  is 

unchanged  and  the  mag- 
nitude of  the  flux  is  varying  between  the  same  limits  and 
with  the  same  frequency.  A  dynamo  armature  has  thus 
also  a  hysteresis  loss  which  can  be  computed  as  above. 

88.  Energy  of  a  Magnetic  Field  in  Iron.  In  the  last  chap- 
ter we  saw  that  when  a  magnetic  field  was  set  up  in  a  sub- 
stance of  constant  permeability,  the  energy  storage  per  cubic 
centimeter  could  be  written  as 


Residual^ 
llagnetit 


H 


B2 

w  =  o —  ergs  per  cubic  centimeter. 

OTTJU 


(1) 


This  expression  held,  by  reason  of  its  derivation,  only  where 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     303 

the  material  was  of  constant  permeability.  We  may  use  it, 
however,  to  find  the  additional  energy  which  must  be  put 
into  the  material  to  change  the  flux  density  by  a  small 
amount,  for  when  we  consider  a  sufficiently  small  change, 
the  variation  of  the  permeability  during  such  change  will 
not  be  great  enough  to  affect  the  results.  Differentiating 
the  above  expression,  we  obtain 

dw=*?.  (2) 

4?r  M 

which  gives  the  increment  of  energy  to  be  added  in  order  to 
change  the  flux  density  by  the  differential  amount  dB. 
In  this  expression,  we  may  insert  the  value 

H-f  (3) 


gvng 


dw  =    -  HdB.  (4) 

47T 


This  expression  does  not  contain  AI  and  hence  may  be  used 
in  cases  where  /z  is  not  a  constant  and  where  the  expression 
above  would  be  in  error,  such  for  instance  as  that  of  a  field 
in  iron.  In  order  to  obtain  the  total  amount  of  energy  per 
cubic  centimeter  of  the  field,  we  must  integrate  this  ex- 
pression, obtaining 


w 


=  j-  I  HdB  ergs  per  cubic  centimeter,  (5) 


where,  of  course,  H  is  a  variable  depending  upon  B  and  not 
a  constant.  The  relation  between  H  and  B  is  given  by  the 
B-H  curves  for  the  material;  that  is,  the  magnetization 
curves  of  the  material  plotted  with  gilberts  per  centimeter 
as  abscissas  and  gausses  as  ordinates. 

Applying  this  equation  to  the  magnetization  curves  shown 
in  Fig.  154,  we  may  find  the  energy  per  cubic  centimeter 
necessary  to  raise  the  flux  density  in  the  iron  to  any  given 
value. 


304      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Thus  to  establish  a  flux  density  BI  in  a  sample  originally 
unmagnetized,  we  must  evaluate  the  integral  where  B  varies 
from  zero  to  BI.  The  value  of  this  integral  is  evidently  the 
area  which  is  shaded  in  the  figure.  The  energy  per  cubic 
centimeter  stored  in  the  iron  when  the  flux  density  of  the 

latter  is  raised  to  the  value 
B 


will  therefore  be  equal 
to  j-  times  the  shaded  area, 

the  result  being  in  ergs  per 
cubic  centimeter. 

We  may  note  in  passing 
that  if  the  permeability  had 
been  constant,  the  magnet- 
ization curve  would  have 
taken  the  form  shown  by 
the  dotted  line.  The  energy 
storage  in  this  case  would 


z?, 


H 


The  energy  put  into  the 
magnetic  field  in  magnetizing  a 
specimen  from  0  to  B  is  repre- 
sented by  the  shaded  area. 


1 


have  been  —  times  the  area  of  the  triangle  instead  of  the 
area  that  is  shaded,  or 


B2 


where  M  is  the  ratio  of  final  B  to  final  H.  On  account  of 
the  curvature  of  the  magnetization  curve,  the  energy  per 
cubic  centimeter  stored  in  the  iron  will  thus  always  be  less 
than  this  value. 

In  a  magnetic  circuit  which  is  made  largely  of  iron  and 
contains  a  small  air  gap,  the  energy  storage  will  usually  be 
almost  entirely  in  the  air  gap,  on  account  of  the  large  value 
of  M  which  reduces  the  expression  above  to  a  low  amount. 

89.  Hysteresis  Loops.  When  a  magnetizing  force  is 
applied  to  iron  which  has  been  previously  entirely  demagnet- 
ized, and  then  removed,  the  iron  will  not  return  to  its  former 
magnetic  state,  but  will  retain  a  certain  amount  of  residual 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     305 

magnetism.  In  fact,  the  magnetic  condition  of  the  iron 
depends  not  only  upon  the  magnetizing  force  applied  but 
also  upon  the  immediate  previous  history  of  the  iron.  In 
electrical  machinery,  however,  we  deal  almost  always  with 
flux  that  is  alternating,  so  that  the  magnetizing  force  is 
applied  first  in  one  direction  and  then  in  the  other  alternately. 
After  a  few  reversals  of  the  magnetizing  force  in  this  man- 
ner, the  iron  will  come  to  a  stable  condition  in  which  it  will 
repeat  a  certain  series  of  values  of  flux  densities.  If  we  then 
plot  the  magnetizing  force  against  the  flux  density,  we  shall 
obtain  what  is  known  as  a  hysteresis  loop  for  the  material. 
Such  a  curve  is  shown  in  Fig.  153.  As  the  magnetizing 
force  is  increased  to  a  maximum  first  in  one  direction  and 
then  in  the  other,  the  flux  density  will  also  alternate,  but  the 
flux  density  when  H  is  decreasing  will  be  larger  than  for 
corresponding  values  of  H  when  it  is  increasing.  On  the 
diagram  is  shown,  as  we  have  already  pointed  out,  the  value 
of  the  residual  magnetism  for  the  material,  which  is  the  flux 
density  remaining  when  H  is  entirely  removed,  and  also  the 
value  of  the  coercive  force,  which  is  the  value  of  H  in  the 
opposite  direction  necessary  to  entirely  remove  the  residual 
magnetism. 

We  have  seen  that  when  iron  is  magnetized  to  a  flux 
density  B,  an  amount  of  energy  is  required  of  value 

w  =  J-  Cms.  (6) 


Therefore,  when  the  value  of  H  is  increased  from  zero  to  its 
maximum  value,  the  flux  density  in  the  meanwhile  changing 
from  the  negative  value  (0-1)  to  its  maximum  value  £max, 
an  amount  of  energy  will  be  supplied  to  the  iron  which  is 

equal  to  -r-  times  the  area  included  between  the  curve  1-2  and 
4» 

the  B  axis.  This  area  is  lightly  cross-hatched.  When  the 
magnetizing  force  is  again  decreased  to  zero,  the  flux  density 
meanwhile  decreasing  from  its  maximum  value  to  the  value 
of  the  residual  magnetism  0-3,  energy  will  be  returned  from 


306      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


the  magnetic  field  to  the  circuit  of  amount  equal  to  —  times 

the  area  between  the  curve  2-3  and  the  B  axis.  This  area 
is  cross-hatched  in  addition  with  dotted  lines.  The  net 
amount  of  work  done  on  the  iron,  that  is,  the  excess  of  the 
energy  put  in  over  that  returned,  is  the  difference  between 

these  two  areas  multiplied  by  — .     This  difference  of  area  is 

the  area  1-2-3  of  half  the  hysteresis  loop.  Exactly  the  same 
procedure  occurs  when  H  increases  to  its  maximum  value 
in  the  opposite  direction  and  again  decreases  to  zero.  For 
a  complete  cycle  in  which  the  flux  density  varies  completely 
from  a  maximum  in  one  direction  to  a  maximum  in  the 
opposite  direction  and  back  again,  the  net  amount  of  work 

supplied  is  thus  equal  to  —  times  the  area  of  the  hysteresis 

loop.     This  energy  is  converted  into  heat. 

The  hysteresis  loss  per  cycle  in  a  sample  of  iron  subjected 

to  an  alternating  magnetizing  force  is  equal  to  j-  times  the 

area    of    the    hysteresis 
loop.     If    H  is   plotted 
in  gilberts  per  centimeter 
and   B   in   gausses,   the 
energy  thus   found  will 
be    in    ergs    per    cubic 
centimeter   of    the    ma- 
terial per  cycle.    If  other 
scales  are  used,   correc- 
tion must  be  made  ac- 
cordingly.      These    cor- 
rections will  appear  from 
the  examples  given  below. 
The  area  of  the  hysteresis  loop  is  thus  a  measure  of  the 
amount  of  the  hysteresis  loss.     When  this  loss  is  small,  the 
area  will  be  long  and  narrow,  as  for  instance  in  Fig.  155, 
which  shows  the  hysteresis  loop  for  a  sample  of  carefully 


FIG.  155.  The  hysteresis  loop  for 
soft  iron.  A  narrow  loop  means  a 
small  loss. 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     307 


H 


annealed  soft  iron.     Fig.  156  shows  the  other  extreme,  a 
hysteresis  loop  for  a  sample  of  hard  steel. 

The  hysteresis  loop  for  a  material  can  be  found  experi- 
mentally by  applying  successive  values  of  magnetizing  force, 
measuring  the  corresponding  values  of  flux  density  and 
plotting  the  results.  In  making  these  measurements  it  is, 
of  course,  necessary  to  be  sure  that  H,  and  hence  the  mag- 
netizing current,  increases  steadily  and  without  decreasing 
until  it  reaches  its  maximum  value,  and  then  decreases 
steadily  to  zero.  For  such  work  a  permeameter  is  especially 
useful.  This  instrument  consists  simply  of  an  arrangement 
whereby  a  measuring  coil  may 
be  made  to  surround  a  sample 
of  iron  to  be  tested,  and  may 
suddenly  be  removed  by  open- 
ing the  magnetic  circuit  and 
pulling  the  coil  out.  It  is 
certain  in  this  way  that  the 
flux  through  the  coil  decreases 
to  zero,  so  that  a  ballistic 
galvanometer  connected  with 
it  shows  a  deflection  corre- 
sponding to  the  flux  at  the 
instant  the  coil  is  removed. 

To  find  the  value  of  B  corresponding  to.  a  given  value  of 
H  on  the  hysteresis  loop,  we  consequently  proceed  as  follows. 
Place  the  measuring  coil  in  position.  Reverse  the  mag- 
netizing current  several  times  to  be  sure  that  the  flux  is 
accurately  following  the  hysteresis  loop  in  value.  Then, 
moving  always  in  the  correct  direction,  increase  the  mag- 
netizing current  until  the  desired  value  of  H  is  present. 
Snap  the  measuring  coil  out  of  position  and  note  the  deflec- 
tion of  the  ballistic  galvanometer.  From  this,  calculate 
the  value  of  B.  Plot  these  values  of  B  and  H  as  a  point 
on  the  hysteresis  loop.  By  proceeding  in  this  manner  the 
entire  hysteresis  loop  can  be  traced. 


FIG.  156.  The  hysteresis  loop 
for  hard  steel.  The  wide  loop 
means  a  large  loss. 


308      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Example  1.  Let  us  examine  an  example  of  the  application  of 
these  principles.  A  certain  transformer  has  a  core  containing  in 
all  500  cubic  inches  of  transformer  iron.  This  is  placed  on  a 
circuit  and  magnetized  by  an  alternating  current  so  that  its 
maximum  flux  density  is  80,000  lines  per  square  inch.  The 
frequency  of  the  magnetizing  current  is  60  cycles  per  second; 
that  is,  the  iron  is  magnetized  in  each  direction  sixty  times 
every  second,  a  complete  cycle  thus  occupying  one-sixtieth  of 
a  second.  A  sample  of  this  iron  is  tested  and  its  hysteresis 
loop  plotted.  This  loop  is  plotted  to  a  scale  such  that  one 
space  in  a  vertical  direction  is  equal  to  1,000  lines  per  square 
inch,  and  one  space  in  a  horizontal  direction  equal  to  one  am- 
pere-turn per  inch  of  length  of  the  core.  The  area  of  this  hys- 
teresis loop  is  measured  by  a  planimeter  and  is  found  to  be 
400  square  spaces.  We  wish  to  compute  the  watts  loss  in 
this  core  due  to  hysteresis. 

In  the  first  place,  one  square  space  is  equal  to  the  product  of 
1000  lines  per  square  inch  by  one  ampere-turn  per  inch.  Chang- 
ing these  values  to  gausses  and  gilberts  per  centimeter,  we  have 

1000  0.47T 

1  square  space  =    — —  gausses  X  ^-rr  gilberts  per  centimeter. 
b.4o  J.54 

The  loss  is  equal  to  JL  times  the  area  of  the  loop  when  con- 

4ir 

verted  to  terms  of  gausses  and  gilberts  per  centimeter.  The 
energy  loss  per  cubic  centimeter  of  the  material  for  each  reversal 
of  the  magnetizing  current  is  therefore 

-  X  400  X  -^-rz  X  J^-TT  ergs  per  cubic  centimeter  per  cycle. 

TtTT  O*^rO  ^*Oi 

Multiplying  this  by  10~7  to  convert  to  joules,  we  have 

4  X  104 

3  X  10~7  joules  per  cycle  per  cubic  centimeter. 

2i  .  •  >  1 

Let  us  multiply  this  energy  by  the  number  of  cubic  centimeters 
in  the  core  and  by  the  number  of  cycles  per  second.  We  shall 
then  find  that  the  total  watts  loss  is 

4  v  104 

-==^-  X  1C-7  X  2.548X  500  X  60  watts, 
2.543 

or  the  total  loss  is 

120  watts. 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     309 

The  hysteresis  loss  for  a  sample  of  iron  in  an  electric  ma- 
chine is  evidently  proportional  to  the  frequency.  If  the 
frequency  is  doubled,  the  hysteresis  loss  will  also  be  doubled, 
other  factors  remaining  the  same.  Hysteresis  loss  is  for 
this  reason  a  matter  of  much  greater  moment  in  high-speed 
machinery  than  in  low-speed  machinery.  In  transformer 
design  also,  it  is  of  greater  importance  with  high  frequency. 
This  is  particularly  true  where  very  high  frequencies  are 
employed.  In  alternating-current  generators  for  very  high 
frequencies,  such  as  the  Alexanderson  alternator  for  the  gen- 
eration of  frequencies  up  to  100,000  cycles  per  second  for  radio 
telegraph  purposes,  the  prevention  of  very  high  hysteresis 
loss  is  a  matter  of  great  moment.  This  is  accomplished  by 
using  an  excellent  grade  of  iron  which  has  a  low  area  of  the 
hysteresis  loop,  and  also  by  utilizing  a  very  small  volume  in- 
deed. These  alternators  are  so  arranged  that  the  iron  part 
in  general  carries  a  constant  flux  with  the  exception  of  a 
small  volume  which  carries  an  alternating  flux.  The  hys- 
teresis loss  per  cubic  centimeter  of  this  material  is  high,  but 
since  a  very  small  volume  is  used,  the  total  losses  are  kept 
within  reason. 

In  considering  core  losses  in  electrical  machinery  where 
the  flux  density  is  varying,  we  must  consider  not  only  the 
hysteresis  loss  as  dealt  with  in  this  chapter,  but  also  eddy- 
current  loss  as  treated  in  the  next  chapter*.  Together  they 
are  grouped  as  core  loss.  Wherever  the  flux  density  in 
electrical  machinery  is  varying  and  iron  is  present,  these 
losses  must  be  taken  into  consideration.  Not  only  will 
they  decrease  the  efficiency  of  the  apparatus  in  question,  but 
it  is  necessary  to  make  provision  for  properly  getting  rid 
of  the  heat  evolved  to  prevent  an  undue  temperature  rise  of 
any  part  of  the  machinery. 

Prob.  1-9.  A  hysteresis  loop  for  a  given  specimen  of  iron 
(a  rod  0.236  inch  in  diameter  and  8.42  inches  long)  has  an 
area  of  8.36  square  inches.  The  loop  is  plotted  on  cross-section 
paper  so  that  an  ordinate  of  one  inch  is  equivalent  to  4000  gaus- 


310      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


ses,  an  abscissa  of  one  inch  to  20  gilberts  per  centimeter.  What 
energy  loss  in  ergs  per  cubic  centimeter  per  cycle  is  represented 
by  the  area  of  the  loop?  Express  this  energy  in  joules  per 
cubic  inch  per  cycle. 

Prob.  2-9.  If  the  hysteresis  loop  of  the  specimen  of  iron 
in  Prob.  1-9  were  taken  with  the  magnetizing  coil  connected 
to  a  25-cycle  source,  what  would  be  the  energy  in  joules  per 
second  or  watts  dissipated  in  the  iron? 

90.  Mean  Magnetization  Curves.  Froelich's  Equation. 
When  an  alternating  magnetizing  force  is  applied  to  a  sample 
of  iron,  we  obtain  a  hysteresis  loop  which  shows  the  vari- 
ation in  flux  density  corresponding  to  the  variation  in 
magnetizing  force.  If  the  maximum  value  of  the  magnet- 
izing force  is  adjusted  to  different  values,  we  obtain  a 

series  of  such  hysteresis 
loops  corresponding  to 
the  different  values  of 
the  maximum  flux  den- 
sity of  each.  Such  a 
series  of  loops  is  shown 
in  Fig.  157.  The  dotted 
curve  which  connects  the 
peaks  of  these  hysteresis 
loops  is  usually  plotted 
as  the  mean  magnetiza- 
tion curve  of  the  spec- 


FIG.  157.    Hysteresis    loops    for    dif- 
ferent maximum  flux  densities. 


imen.  The  curves  be- 
tween ampere-turns  and 
flux  densities  which  are 

shown  in  Fig.  61  of  Chapter  VI  are  obtained  in  this  manner. 
This  mean  magnetization  curve  is  usually  the  curve  in 
which  we  are  interested  in  determining  the  relation  between 
B  and  H,  for  we  usually  deal  with  periodically  varying 
magnetizing  forces. 

It  was  determined  by  Froelich  that  the  magnetization 
curve  of  irons  and  steels  may  be  approximately  represented 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     311 
by  an  equation  of  the  form 


. 

This  is  the  equation  of  a  hyperbola.  It  is  an  empirical 
equation  since  it  is  derived  from  the  study  of  a  mass  of  data. 
It  does  not  apply  to  the  extreme  lower  end  of  the  curves, 
but  fortunately  is  a  very  close  approximation  for  that  part 
of  the  magnetization  curve  which  is  in  greatest  use. 

The  value  of  the  constants  in  Froelich's  equation  for  any 
particular  magnetization  curve  may  be  derived  as  follows. 
If  .  aH 


Let  6  =  J  (10) 

and  -  =  K.  (11) 

The  equation  becomes 

-|  =  K  (J  +  H)  .  (12) 

This  is  the  equation  for  a  straight  line  with  intercept  J  and 

TT 

slope  K  if  ^-  is  considered  one  of  the  variables.     We  have 
r> 

TT 

only  to  determine  sufficient  values  of  -5-  (which  is  the  re- 

£>   • 

luctivity  of  the  material)  and  plot  these  values  against  H. 
The  result  is  a  straight  line  as  in  Fig.  158  if  the  curve  fol- 
lows Froelich's  equation. 

The  straight  line  in  Fig.   158  is  the  plot  between  the 

TT 

values  of  -=  and  H  taken  from  the  magnetization  curve  for 

D 

annealed  sheet  steel  which  is  drawn  on  the  same  sheet. 
The  equation  for  this  straight  line  is 

|  =  K  (J  +  H), 

when  J  is  the  intercept  on  the  H  axis  and  K  is  the  slope  of 
the  curve. 


312      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


\ 


III! 


X 


rt\ 


\ 


s; 


«n 

*g  8 

S    o 


s-8 


•^  a 


£ 

| 

00 

H 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     313 
But  from  equations  (10)  and  (11) 


a 
and 

J  =  6. 

The  value  for  a  in  Froelich's  equation  is  therefore  the 

TT 

reciprocal  of  the  slope  of  the  curve  between  -=  and  H,  and 

£> 

the  value  for  b  is  the  intercept  on  the  axis  of  H. 

TT 

Note  that  in  plotting  the  curve  between  -5  and  H  the 

£> 

extreme  low  values  of  H  should  not  be  used.* 

Prob.  3-9.  Determine  Froelich's  equation  for  the  mag- 
netization curve  of  annealed  sheet  steel  as  shown  in  Fig.  158. 

Prob.  4-9.  What  part  of  the  curve,  if  any,  for  cast  iron  as 
given  in  Fig.  61  satisfactorily  follows  Froelich's  equation? 
Determine  the  constants  a  and  b  for  this  part  of  the  curve. 

91.  Methods  of  Demagnetizing  Steel.  If  we  start  out 
with  an  entirely  unmagnetized  piece  of  iron  and  gradually 
increase  the  magnetizing  force,  a  variation  of  flux  density 
will  be  obtained  which  will  give  a  curve  very  similar  to  the 
dotted  curve  of  Fig.  157.  For  most  purposes  the  curve 
thus  obtained  and  that  of  Fig.  158  are  identical.  The  mean 
curve  obtained  as  the  locus  of  the  peaks  of  the  successive 
hysteresis  loops  is,  however,  the  one  generally  employed  in 
practice.  Of  course  it  must  be  realized  that  in  most  of  the 
iron  with  which  we  deal  in  electrical  apparatus,  the  hysteresis 
loop  will  be  long  and  narrow,  more  of  the  appearance  of  Fig. 
155  rather  than  the  exaggerated  form  which  we  have  shown 
in  Fig.  157  for  convenience.  In  other  words,  the  flux  density 
on  rising  and  decreasing  magnetizations  will  not  differ  as 
much  as  here  shown,  when  we  are  dealing  with  steel  which  is 
of  proper  quality  to  be  used  in  electrical  apparatus  where 
alternating  magnetizing  forces  are  employed. 

*  For  precise  methods  of  obtaining  values  of  J  and  K  from  curves, 
see  Lipka's  "  Graphical  and  Mechanical  Computation." 


314      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

When  the  magnetizing  force  is  removed  from  a  sample  of 
iron,  there  will  remain  a  certain  flux  density,  which  we  have 
said  is  called  the  residual  magnetism.  In  order  to  com- 
pletely demagnetize  the  iron  it  is  necessary  to  employ  a 
coercive  force,  which  is  a  magnetizing  force  in  the  opposite 
direction,  and  of  sufficient  amount  to  neutralize  this  residual 
magnetism.  There  is  another  and  more  convenient  method, 
however,  of  demagnetizing  a  sample  of  iron.  If  the  magnet- 
izing force  is  made  alternating  and  then  is  gradually  de- 
creased to  zero,  the  flux  density  will  vary  in  accordance  with 
a  hysteresis  loop  of  constantly  decreasing  size,  which  will 
finally  wind  its  way  into  a  very  small  loop  about  the  origin. 
By  thus  varying  the  magnetizing  force  and  at  the  same  time 
decreasing  it  to  zero,  the  flux  can  be  removed  almost  en- 
tirely from  even  a  sample  of  hard  steel.  Thus  to  demagnet- 
ize the  hair-spring  of  a  watch  we  may  place  it  near  an  open- 
core  solenoid,  the  winding  of  which  is  carrying  an  alternating 
current,  and  with  the  current  still  turned  on  in  the  solenoid 
gradually  remove  the  watch  from  the  field.  An  almost 
equal  effect  can  be  obtained  by  bringing  the  watch  close  to  a 
permanent  magnet,  rotating  it  rapidly,  and  gradually  with- 
drawing it  from  the  field.  In  either  of  these  cases  the  mag- 
netizing force  applied  to  the  spring  is  alternated  and  gradu- 
ally decreased  to  zero. 

92.  The  Steinmetz  Equation.  We  have  seen  that  the  hys- 
teresis loss  which  is  produced  when  the  magnetizing  force 
applied  to  a  given  sample  of  iron  is  alternated  is  proportional 
to  the  volume  of  the  iron  and  also  to  the  frequency  of  al- 
ternation. This  means  that  hysteresis  loss  per  cubic  centi- 
meter per  cycle  is  a  constant  for  a  given  maximum  flux 
density.  It  is  of  great  practical  interest  to  know  how  this 
loss  varies  with  the  maximum  flux  density  employed. 

There  is  no  theoretical  law  governing  this  variation,  but 
Steinmetz,  as  the  result  of  an  extended  series  of  tests,  has 
derived  an  empirical  equation  which  fits  the  facts  sufficiently 
closely  for  engineering  purposes  if  used  over  a  moderate 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     315 

range.     This  equation  states  that  the  hysteresis  loss  varies 
as  the  1.6th  power  of  the  maximum  flux  density.     That  is, 

w  —  17 /  BmaxL6  ergs  per  cubic  centimeter  per  second,    (13) 
where 

w  =  loss  in  ergs  per  second  per  cubic  centimeter  of  the 

material, 
#max  =  maximum  flux  density, 

/  =  number  of  cycles  per  second  of  the  magnetizing 
force. 

The  flux  density  used  is  the  maximum  flux  density  ob- 
tained during  the  cycle  and  is  expressed  in  gausses.  To 
obtain  the  loss  in  watts  per  cubic  centimeter,  we  simply 
multiply  the  value  obtained  from  the  above  equation  by 
10~7.  Eta  (77)  is  a  coefficient,  called  the  coefficient  of  hys- 
teresis loss,  which  depends  upon  the  kind  of  iron  being  used. 
For  good  silicon  steel,  the  value  of  rj  is  0.001.*  The  value 
for  soft  iron  is  0.002  to  0.004.  For  hard  cast  steel  it  is  as 
high  as  0.025,  and  in  an  extreme  case  for  tungsten  steel  it 
may  run  as  high  as  0.058. 

It  should  be  emphasized  that  this  formula  is  an  em- 
pirical formula  only  and  simply  sums  up  the  results  of  ex- 
perience. It  applies  fairly  well  for  the  ordinary  flux  den- 
sities used  in  practice,  that  is,  from  1500 -to  12,000  gausses. 
If  the  flux  density  is  higher  than  this,  the  loss  will  be  found 
to  increase  faster  than  is  indicated  by  the  formula.  For 
very  low  values  of  flux  density,  also,  the  formula  will  be 
found  to  be  greatly  in  error. 

As  an  example  of  the  use  of  this  formula,  suppose  that  the 
hysteresis  loss  in  a  certain  transformer  is  400  watts.  It  is 
proposed  to  double  the  voltage  applied  to  this  transformer, 
which  will  result  in  practically  a  doubling  of  the  maximum 

*Note:  Lloyd,  "  Magnetic  Hysteresis,"  Journal  of  the  Franklin 
Institute,  July,  1910, 


316      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


flux  density.     If  the  frequency  is  maintained  as  before, 
what  will  be  the  new  hysteresis  loss? 

Since  the  loss  varies  as  the  1.6th  power  of  the  flux  density, 
doubling  the  flux  density  will  make  the  new  loss 


or 


400  X  21-6  watts, 
1210  watts. 


The  Steinmetz  formula  is  based  on  the  hysteresis  loop. 
It  states  in  effect  that  the  area  of  this  hysteresis  loop  varies 
as  the  1.6th  power  of  its  height.  The  formula  applies  only 
for  normal  hysteresis  loops;  that  is,  in  cases  where  the  maxi- 
mum flux  density  attained  in  one  direction  is  equal  to  the 

maximum  flux  density 
obtained  in  the  op- 
posite direction.  For 
unsymmetrical  loops, 
such  as  would  be  ob- 
Time  tained  where  an  alter- 

nating and  a  constant 
magnetizing  force  are 
applied  simultaneously, 
the  formula  will  not 
hold.  It  will  not  hold 
either  for  cases  where 
the  magnetizing  force  is 
not  increased  contin- 
uously to  its  maximum 
value.  Suppose,  for  instance,  that  we  apply  to  a  mag- 
netizing coil  a  current  which  varies  in  accordance  with  the 
curve  shown  in  Fig.  159.  It  will  be  noted  that  the  current 
increases  for  a  while,  then  decreases  slightly  and  finally 
increases  to  its  maximum  value,  then  decreases  to  zero, 
and  then  repeats  its  performance  in  the  opposite  direction. 
The  result  will  be  a  hysteresis  loop  with  a  re-entrant  loop  as 
shown  in  Fig.  160.  The  small  twist  in  the  curve  is  caused 


FIG.  159.  A  non-sinusoidal  exciting 
current  of  this  form  produces  a  hys- 
teresis loop  of  the  form  shown  in 
Fig.  160. 


B 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     317 

by  the  decrease  in  the  current  before  it  has  arrived  at  its 
maximum  value. 

For  such  cases  of  distorted  wave  forms  as  this,  the  formula 
will  naturally  not  apply  even  approximately.  Such  cases 
can  be  analysed  only 
by  actually  plotting  the 
hysteresis  loop  itself 
and  measuring  the  area. 
In  measuring  this  area 
with  a  planimeter, 
we  pass  around  the 
curve  as  it  is  actually 
traced.  This  process 

will  add  in  the  area  of 

,1  ni  .        ,  ,.        FIG.    160.      The  form  of  hysteresis  loop 

the  small  loops  m  addi-        produoed  by  a  current  of  the  form 

tion  to  the  area  of  the        fn  Fig.  159. 

large  loop.     The  result 

obtained  for  the  area  will  be  the  actual  hysteresis  loss  under 

the  conditions  which  produce  such  a  distorted  curve. 

Prob.  6-9.  The  sheet-steel  core  of  a  transformer  has  a 
mean  length  of  32  inches  and  a  cross-section  of  2  square  inches. 
When  110  volts,  60  cycles,  is  impressed  on  the  winding,  the 
maximum  flux  density  is  40,200  lines  per  square  inch.  For 
this  particular  steel  77  =  0.0013.  Find  the  hysteresis  loss  in 
watts  per  cubic  inch.  Find  it  in  watts  per  pound,  assuming 
the  volume  of  iron  to  be  0.9  times  the  volume  of  the  core,  owing 
to  scale  on  the  laminations  of  which  the  core  is  made  up.  What 
is  the  total  hysteresis  loss  in  the  core? 

Prob.  6-9.  A  transformer  core  with  a  volume  of  8.56  cubic 
inches  has  a  hysteresis  loss  at  25  cycles  of  2.14  watts.  The 
maximum  flux  density  in  the  material  is  60,000  lines  to  the 
square  inch.  Find  77.  If  the  frequency  is  raised  to  60  cycles 
and  the  flux  density  reduced  one-half,  what  will  be  the  new 
loss  in  watts  per  cubic  centimeter? 

93.  Effect  of  the  Composition  of  Steel.  It  has  been 
noted  above  that  the  chemical  composition  of  a  steel  has  a 
very  great  effect  indeed  upon  its  hysteresis  loss.  A  hard 


318      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

tungsten  steel  may  have  fifty  times  as  much  hysteresis  loss 
for  a  given  flux  density  as  a  silicon  steel  of  proper  composition. 

The  hysteresis  coefficient  and  the  permeability  depend 
not  only  upon  the  chemical  composition  of  the  steel  but  also 
upon  its  heat  treatment.  Carefully  annealing  a  sample  of 
iron  may  reduce  its  hysteresis  loss  by  as  much  as  fifty  per- 
cent. The  properties  also  vary  with  the  temperature. 
Both  iron  and  steel,  in  fact,  lose  all  their  magnetic  qualities 
at  about  750  degrees  centigrade  and  have  a  permeability  of 
practically  unity  above  that  temperature.  This  is  called 
the  recalescence  point  of  the  material. 

If  ordinary  steels  are  subjected  to  a  temperature  of  about 
100  degrees  centigrade  for  a  long  period,  the  hysteresis  co- 
efficient will  gradually  increase.  This  process  is  known  as 
aging.  Modern  silicon  steels,  such  as  are  used  in  transformer 
construction,  are  almost  entirely  free  from  this  aging.  This 
and  a  fairly  low  hysteresis  coefficient  are  the  reasons  for 
the  use  of  such  steel  in  cases  where  alternating  magnetizing 
forces  are  employed.  These  steels  usually  contain  from  three 
to  four  percent  silicon.  It  is  necessary  that  the  carbon 
content  be  low,  less  than  0.1  percent.  A  small  percentage 
of  either  sulphur  or  phosphorus  will  render  the  steel  ex- 
tremely poor  electrically.  The  sheets,  after  rolling,  are  an- 
nealed at  about  the  recalescence  point  and  cooled  very 
slowly. 

In  the  study  of  eddy-current  loss,  we  shall  see  later  that 
high  resistivity  is  desirable  in  transformer  steels.  This 
property  is  also  obtained  in  silicon  sheet  steel. 

For  low  retentivity,  a  Norway  iron  is  usually  employed. 
This  material  has  a  very  high  permeability  but  a  low  value 
of  residual  magnetism,  and  a  low  coercive  force.  It  is  ac- 
cordingly used  for  the  armatures  of  electromagnets  or  relays 
in  places  where  an  attraction  due  to  residual  magnetism 
is  undesirable.  Ordinary  wrought  iron  has  a  high  per- 
meability but  its  hysteresis  loss  is  fairly  large.  Cast  iron 
has  a  relatively  low  permeability,  some  samples  having  a 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     319 

maximum  of  only  150,  as  compared  with  values  of  3000  for 
wrought  iron.  Malleable  cast  iron  may  have  a  permeability 
as  high  as  700. 

There  has  recently  been  produced  an  electrolytic  iron 
melted  in  a  vacuum  which  has  extraordinary  properties.* 
A  maximum  permeability  of  19,000  has  been  obtained  at  a 
density  of  9500  gausses.  The  hysteresis  loss  was  half  that 
found  for  the  best  grade  of  commercial  transformer  iron. 
Unfortunately,  however,  at  the  present  time  such  iron  is 
very  expensive  and  is  simply  a  laboratory  product. 

94.  Permanent  Magnets.  In  electrical  apparatus  where 
only  a  moderate  field  strength  is  required,  a  permanent 
magnet  is  often  employed.  A  familiar  example  is  the  mag- 
net employed  to  supply  flux  to  a  magneto  used  for  gas-engine 
ignition.  For  dynamos  of  large  output,  an  electromagnet 
is  always  used,  but  where  the  output  is  very  small,  as  in  the 
above  case,  a  permanent  magnet  can  sometimes  be  em- 
ployed. In  electric  meters  also  permanent  magnets  are 
used,  and  in  this  case  they  must  be  strictly  constant  in  field 
strength  over  long  periods  of  time  in  order  that  the  meters 
may  not  show  large  errors.  Large  numbers  of  permanent 
magnets  are  also  used  on  integrating  wattmeters,  such  as  we 
find  on  house  circuits  for  measuring  the  kilowatt-hours 
used  on  alternating-current  lighting  circuits. 

Such  magnets  are  usually  constructed  of  tungsten  steel, 
containing  about  seven  percent  of  tungsten.  This  steel  is 
hardened  carefully  by  heating  it  to  the  recalescence  point 
and  quenching.  This  must  be  done  with  great  care.  If 
the  temperature  is  incorrect  by  even  fifteen  or  twenty  de- 
grees, a  considerable  reduction  in  the  retentivity  will  be 
found.  It  is  also  necessary  to  quench  very  carefully  in  order 
to  avoid  cracking  the  steel. 

The  magnets  are  then  magnetized  by  surrounding  them 
with  a  coil  through  which  can  be  passed  a  heavy  current, 
usually  from  a  storage  battery.  The  magnetizing  force 

*  Proceedings  A.  I.  E.  E.,  February,  1915,  p.  236. 


320      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

should  be  such  as  to  saturate  the  steel,  magnetizing  forces 
of  100  ampere-turns  per  centimeter  or  so  being  employed. 
Horseshoe  magnets  are  usually  magnetized  in  pairs  in  order 
to  give  them  a  complete  magnetic  circuit.  Magnets  for 
magnetos  generally  have  iron  keepers  placed  across  the  ends 
at  the  time  when  they  are  magnetized.  In  very  careful 
magneto  construction,  the  keeper  is  never  removed  com- 
pletely until  the  magnet  is  placed  in  position,  when  the  iron 
parts  of  the  magneto  take  its  place.  A  certain  increased 
magnetization  will  usually  result  from  jarring  a  magnet 
somewhat  while  the  magnetizing  force  is  applied.  This 
apparently  allows  the  molecules  to  move  more  freely  into  the 
magnetic  position.  By  using  care  with  good  steel,  a  per- 
manent magnet  may  be  constructed  which  will  give  in  an 
air  gap  of  small  length  a  flux  density  of  about  6000  gausses. 

If  a  freshly  charged  magnet  is  violently  jarred,  it  will  lose 
some  of  its  magnetism.  Snapping  the  keeper  into  place  is 
especially  damaging.  In  the  preparation  of  magnets  which 
must  maintain  an  extremely  constant  flux,  they  are  sub- 
jected to  artificial  jarring  after  being  magnetized  to  remove 
such  parts  of  the  magnetism  as  are  likely  to  disappear  in  time. 

Permanent  magnets,  when  left  standing,  will  gradually 
lose  a  portion  of  their  magnetism,  and  after  a  time  settle 
down  to  constant  values,  unless  they  are  abused  in  some 
way.  Meter  magnets  are  often  artificially  aged  by  main- 
taining them  at  a  temperature  of  100  degrees  centigrade  for 
several  days  in  order  to  arrive  quickly  at  this  final  value  of 
strength.  They  then  become  very  permanent  and  will 
not  change  their  strength  appreciably  in  a  long  time. 


SUMMARY    OF   CHAPTER  IX 

THE  RESIDUAL  MAGNETISM  is  the  magnetism  remain- 
ing in  a  material  after  the  magnetizing  force  has  been  re- 
moved. 

MAGNETIC  HYSTERESIS  LOSS  is  the  heat  loss  in  mag- 
netic materials  due  to  the  LAGGING  of  the  values  of  flux 
density  behind  those  of  the  magnetizing  force. 

A  HYSTERESIS  LOOP  is  formed  if  the  values  of  flux  den- 
sity are  plotted  against  the  corresponding  values  for  magnet- 
izing force  throughout  a  complete  cycle  of  positive  and  negative 
values  of  magnetizing  force.  The  area  of  this  loop  is  4ir 
times  the  hysteresis  loss  per  cycle  per  cubic  centimeter  of  the 
specimen. 

THE  ENERGY  OF  A  MAGNETIC  FIELD  IN  IRON  is 
expressed  by  the  equation 

w  =  —  I  HdB  ergs  per  cubic  centimeter, 
47rJ 

where  H  is  a  variable  depending  upon  the  value  of  B. 
The  STEINMETZ  EQUATION  for  hysteresis  loss  is 


w  =  77  f  B1'6    ergs  per  cubic  centimeter  per  sec., 


where 


97  is  the  hysteresis  coefficient  depending  upon  the  material, 
f  is  the  number  of  cycles  per  second  of  the  magnetizing 
force. 

FROELICH'S  EQUATION  applies  approximately  to  all  but 
the  lower  and  upper  ends  of  the  MEAN  magnetization  curve 
of  magnetic  materials. 

B. 


b  +H 

where  a  and  b  are  constants  depending  upon  the  material, 

321 


322      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

To  determine  the  values  of  these  constants,  plot  H  against 
H/B  and  call  J  the  intercept  on  the  x  axis  and  K  the  slope  of 
the  curve,  which  will  be  a  straight  line  throughout  the  region 
where  the  magnetization  curve  follows  Froelich's  equation. 
Then 

1 

a=K 
and 

b  =  J. 

TO  DEMAGNETIZE  A  MAGNETIC  MATERIAL,  place  it 
in  an  alternating  field  and  gradually  decrease  the  strength  of 
the  field. 

THE  COMPOSITION  AND  HEAT  TREATMENT  of  mag- 
netic materials  have  great  effect  upon  the  hysteresis  coefficient 
of  the  materials. 

THE  BEST  PERMANENT  MAGNETS  are  made  of  tung- 
sten steel.  When  properly  aged  and  carefully  handled,  such 
a  magnet  will  retain  a  practically  constant  flux  density  for 
years.  Heat  and  rough  handling  will  materially  decrease  its 
strength. 


PROBLEMS    ON   CHAPTER   IX 

Prob.  7-9.  The  following  data  were  taken  in  the  Technical 
Electrical  Measurements  Laboratory,  Mass.  Inst.  of  Tech.,  on 
a  bar  of  iron  by  means  of  a  "  Koepsel  Permeameter,"  the 
readings  being  taken  successively  in  this  order: 

H  gilberts /centimeter  B  gausses 

+    0  +0 

+  20  +  5000 

+    4.5  +  4000 

+    0  +3400 

-  6.0  +1750 

-  10.0  -  100 
-14.5  -2500 
-18.5  -4200 
-20.5  -5000 

-  5.5  -  4000 

-  2.0  -3500 
+    0  -  3200 
+    7  -    750 
+  10.5  +1600 
+  15.0  +3600 
+  20  +.5000 

Plot  the  values  and  determine  from  the  hysteresis  loop  the 
coercive  force  and  the  relation  between  the  residual  magnetism 
and  the  maximum  flux  density.  Note  that  this  is  not  the 
retentivity  since  the  maximum  flux  density  does  not  represent 
saturation.  The  loop  formed  may  be  assumed  to  represent 
with  sufficient  accuracy  conditions  after  many  flux  reversals. 

Prob.  8-9.  With  the  data  of  Prob.  7-9  find  the  energy  in 
ergs  per  cycle  represented  by  the  hysteresis  loop. 

Prob.  9-9.  The  same  specimen  of  iron  as  in  Prob.  7-9  is 
tested  to  a  higher  flux  density  by  the  same  method  with  the 
following  result: 

323 


324      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

H  gilberts /centimeter  B  gausses 

+  66.5  +12000 

+  41.0  +11250 

+  19.0  +10000 

+  9.0  +  9000 

+  5.5  +  8600 

+  3.0  +  8200 

0  +  7600 

-  2  +  7000 

-  4  +  6500 

-  7  +  5250 

-  10  +  3600 

-  15  -   250 

-  19  -  2800 
-20.5  -  4000 
-27.5  -  6900 

-  34  -  8500 

-  44  -  10250 

-  66  -  12000 

-  32  -  11100 

-  22.5  -  10500 

-  10.0  -  9200 

0  -    7800 

+  10  -    3200 

+  15  +      100 

+  19  +    3100 

+  25  +    5800 

+  35  +    8400 

+  46  +  10000 

+  66.5  +12000 

From  a  plot  of  these  data  to  the  same  scale  as  that  used  in  Prob. 
7-8,  determine  the  coercive  force  and  the  residual  magnetism. 
Superimpose  this  plot  on  a  plot  of  the  data  of  Prob.  7-8,  and 
sketch  the  mean  magnetization  curve  as  determined  by  the 
points  of  the  two  hysteresis  loops. 

Prob.  10-9.  The  coefficient  of  hysteresis  loss,  77,  determined 
from  the  above  data  is  0.00192  and  the  exponent  of  £max. 
1.56,  which  agrees  experimentally  with  the  exponent  1.6  as 
determined  by  Steinmetz.  What  would  be  the  watts  loss  per 
pound  for  this  specimen  at  60  cycles? 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL     325 


Prob.  11-9.     The  data  for  a 

magnetization  curve  are  as  foL 

lows  in  c.  g.  s.  units: 

H 

B 

5.5 

550 

12 

1776 

19 

4500 

25 

7100 

40 

9700 

66 

12075 

109 

13875 

Plot  a  curve  such  as  Fig.  154  and  determine  the  energy  in  ergs 
per  cubic  centimeter  stored  in  the  iron  when  the  flux  density 
is  77,000  lines  per  square  inch.  What  is  the  percentage  differ- 
ence between  this  energy  and  the  amount  of  energy  which  would 
have  been  stored  had  the  permeability  had  the  same  constant 
value  that  it  had  at  the  density  of  77,000  lines  to  the  square  inch? 
(Assume  that  Simpson's  rule  will  apply  to  the  curve.) 

Prob.  12-9.  The  following  design  constants  for  a  20-kilowatt, 
220-2200-volt  transformer  are  representative  of  American 
practice : 

Mean  length  of  core,  43  inches, 

Cross-section  of  core,  4.6  X  4.6  inches, 

Maximum  flux  density,  68  kilolines  to  the  square  inch, 

Material  of  core,  silicon  steel,  17  =  0.0011, 

Net  volume  of  iron,  0.92  times  the  gross  volume  of  core 
as  determined  from  the  dimensions  given. 

Determine  the  hysteresis  loss  in  the  core,  expressing  it  in  joules 
per  cycle.  At  a  frequency  of  60  cycles,  what  is  the  total  hys- 
teresis loss  in  watts? 

Prob.  13-9.     The  following  magnetic-circuit  dimensions  are 
taken  from  a  Weston  5-15-150  voltmeter  (Fig.  185): 
Mean  length  of  permanent  magnet,  12.10  inches, 
Cross-section  (uniform),  1.25  X  0.312  inch, 
Cross-section  of  air  gap  (very  nearly),  1.32  square  inches, 
Length  of  air  gap  (taken  radially),  0.15  inch. 

It  was  proposed  by  Hookham  (Philosophical  Magazine,  1889) 
that  if: 

A  =  area  of  cross-section  of  air  space, 
L  =  distance  between  pole  pieces  (if  two  air  gaps,  L  = 
two  times  the  length  of  one), 


326      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

a  =  cross-section  of  magnet, 

/  =  length  of  magnet  (mean  length  of  lines  of  force  in 
the  permanent  magnet), 

then  for  permanency,  the  ratio 


T 

L    .  a 

should  be  greater  than  100.  This  ratio  is  known  as  the  per- 
manency factor.  Determine  the  permanency  factor  for  the 
magnet  above;  also  the  flux  density  in  the  air  gap  and  in  the 
iron  if  the  flux  crossing  the  gap  is  8.51  X  103  maxwells. 

Prob.  14-9.  A  direct-current  relay  has  a  resistance  of  4 
ohms  and  is  operated  by  a  6-volt  storage  battery.  The  arma- 
ture of  this  relay  starts  to  pick  up  for  a  flux  of  20,000  maxwells. 
Since  the  magnetic  circuit  is  made  of  iron,  the  flux  is  not  pro- 
portional to  the  current.  An  empirical  relation  between  flux 
and  current  which  is  reasonably  accurate  is  as  follows: 

80  Ni 


1  +  0.002  Ni 

where  N  is  the  number  of  turns  and  in  this  case  is  equal  to 
500.  Assume  that  every  flux  line  links  all  of  the  turns.  What 
time  after  the  relay  is  connected  to  the  battery  will  elapse 
before  the  armature  starts  to  pick  up?  The  internal  e.m.f .  of  the 
battery  is  6  volts,  the  internal  resistance  0.01  ohm  and  the 
resistance  of  the  leads  from  relay  to  battery  is  0.19  ohm. 

Prob.  15-9.  In  60-cycle  transformers  with  silicon-steel  cores, 
containing  3  to  4  percent  silicon,  the  hysteresis  loss  is  from 
0.54  to  0.82  watt  per  pound  of  core.*  Using  the  value  of  rj 
in  the  Steinmetz  equation  as  0.0007,  compute  the  maximum 
flux  density  at  which  the  steel  is  worked  when  the  loss  is  0.54 
watt  per  pound.  The  specific  gravity  of  steel  is  7.5. 

Prob.  16-9.  Compute  the  limiting  values  of  77  in  the  Stein- 
metz equation  for  ordinary  annealed  steels  which  have  hys- 
teresis losses  of  from  1.0  to  2.0  watts  per  pound  at  a  flux  density 
of  64,500  lines  per  square  inch  and  at  a  frequency  of  60  cycles.* 

Prob.  17-9.  A  220- volt  relay  has  a  straight  cylindrical  iron 
core,  the  length  and  diameter  of  which  are  respectively  2j 
inches  and  f  inches.  It  is  wound  with  fifteen  hundred  turns, 

*  Trans.  A.  I.  E.  E.,  vol.  XXVIII,  page  465. 


MAGNETIC  PROPERTIES  OF  IRON  AND  STEEL      327 


and  the  resistance  of  the  winding  is  200  ohms.  It  closes  when  the 
flux  density  in  the  core  is  20,000  lines  per  square  inch.  Use  the 
permeability  obtained  by  taking  the  slope  of  the  line  drawn  through 
the  origin  and  the  point  of  maximum  flux  density  on  the  magnet- 
ization curve,  and  determine  the  time  required  for  the  relay  to 
act  after  the  closing  of  the  controlling  key. 

Prob.  18-9.  The  inductance  of  a  circuit  may  be  defined  a3 
the  rate  of  change  of  flux  linkages  with  current.  Bearing  this  in 
mind,  devise  a  graphical  method  for  the  solution  of  the  Prob.  17-9 
taking  into  consideration  the  non-linear  magnetization  curve. 

Prob.  19-9.  The  following  magnetization  data  were  taken  of  a 
3j-h-p.  motor  at  M.I.T.  Between  what  limits  of  flux  density  does 
Froelich's  equation  hold  for  this  motor?  Compute  the  values  of 
a  and  b  in  Froelich's  equation  for  this  part  of  the  curve. 


Field  Current 
in  Amperes 

Generated 
Voltage 
E  =  kB 

Field  Current 
in  Amperes 

Generated 
Voltage 
E  =  kB 

0.10 

15.0 

1.20 

167.0 

0.20 

34.0 

1.4*0 

179.0 

0.30 

52.3 

1.60 

190.0 

0.50 

89.0 

1.80 

199.0 

0.70 

119.5 

2.00 

205.5 

0.80 

131.0 

2.20 

210.2 

0.90 

143.0 

2.40 

214.3 

1.00 

152.0 

2.60 

218.3 

2.80 

221.0 

CHAPTER  X 
GENERATED  VOLTAGES 

We  have  seen  how  by  means  of  a  change  in  the  magnetic 
linkages  an  electromotive  force  is  induced  in  the  turns  of  a 
stationary  coil  or  in  the  conductors  of  a  transmission  line. 
In  this  chapter  will  be  shown  how  this  change  in  magnetic 
linkages,  in  direct-  and  alternating-current  generators,  is 
utilized  to  produce  the  standard  voltages  for  lighting  and 
power. 

95.  Change  of  Linkages.  Wherever  the  magnetic  flux 
linkages  through  a  coil  change  in  number,  there  is  voltage 
induced.  This  voltage  is  proportional  to  the  rate  of  change 
of  linkages,  that  is, 


e=  #      10-8  volts  >  (1) 

where 

N  =  number  of  turns, 
<f>  =  flux  in  maxwells, 
e  =  electromotive  force  in  volts. 

In  all  the  cases  we  have  so  far  considered,  the  coil  has  been 
stationary  and  the  variation  of  flux  through  it  has  been  due 
to  an  increase  or  a  decrease  in  the  amount  of  flux  in  the 
magnetic  circuit.  There  are  other  ways  of  changing  the 
amount  of  flux  in  a  coil,  however.  We  may  move  the  coil 
and  the  magnetic  circuit  with  respect  to  each  other.  It 
makes  no  difference  whether  we  move  the  coil  or  move  the 
flux  as  long  as  the  motion  causes  a  change  in  the  number 
of  flux  lines  threading  the  coil.  In  either  case  there  will  be 
a  change  of  linkages  and  hence  a  voltage  of  the  above 
amount  produced. 


GENERATED   VOLTAGES 


329 


Fig.  161  shows  a  coil  situated  in  an  air  gap  between  two 
heavy  pole  faces.  The  plane  of  the  coil  is  perpendicular  to 
the  flux  lines.  If  the  area  of  the  coil  is  A  and  the  flux 
density  B,  there  is  the  total  number  of  linkages  NBA,  where 
N  is  the  number  of  turns  in  the  coil  (shown  as  one  in  the 
figure  for  convenience). 


FIG.  161.     The  plane  of  the  coil  is  perpendicular  to  the  direction 
of  the  magnetic  flux. 

Suppose  the  coil  is  now  turned  to  the  position  of  Fig. 
162,  that  is,  so  that  its  plane  is  parallel  to  the  flux  lines. 
There  is  now  no  flux  linking  the  coil.  If  T  seconds  are 


FIG.  162.     The  coil  has  been  turned  so  that  its  plane  is  now  parallel 
to  the  magnetic  flux. 

used  in  turning  the  coil  from  one  position  to  the  other,  then 
the  average  rate  of  change  of  linkages  is 


(2) 


(3) 


T  T 

and  the  average  voltage  produced  during  the  process  is 

NBA, 
eM  =  — =—  10~8  volts. 


330      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

An  arrangement  in  which  a  coil  of  wire  is  moved  relative 
to  a  magnetic  field  in  order  to  produce  a  voltage  is  called  an 
electric  generator. 

There  is  another  way  of  looking  at  generation  of  voltage 
in  this  manner  which  is  more  convenient  in  the  study  of 
certain  problems.  In  Fig.  163  is  shown  the  pole  of  a  mag- 
net with  the  flux  lines  issuing  from  it  perpendicularly.  Hori- 
zontally in  this  field  is  located  a  U-shaped  piece  of  wire, 
with  another  short  piece  of  wire  laid  across  to  complete  a 
circuit.  Suppose  that  this  short  piece  of  wire  is  main- 
tained in  contact  with  the  U,  and  moved  bodily  toward  the 


N 


FIG.  163.     The  slider  moves  perpendicular  to  the  lines  of  force  and 
cuts  them. 

right.  The  number  of  lines  of  force  passing  through  the 
closed  electric  circuit  is  then  increasing,  and  accordingly 
there  will  be  a  voltage  set  up.  This  voltage  will  be  in  such 
a  direction  as  to  tend  to  pass  a  current  around  the  loop  in  a 
direction  opposing  the  increase  of  flux,  that  is,  clockwise 
about  the  loop  as  viewed  from  above.  The  direction  of  the 
voltage  produced  is  thus  from  back  to  front  in  the  short 
piece  of  wire. 

It  is  obvious  that  in  the  arrangement  shown  in  Fig.  163, 
the  voltage,  which  is  produced  by  the  motion  of  the  short 


GENERATED  VOLTAGES  331 

cross  piece  of  wire,  must  be  produced  in  this  cross  piece 
only;  for  the  remainder  of  the  apparatus  is  stationary  and 
the  flux  is  constant.  It  is  customary  to  say  that  the  cross 
piece  has  a  voltage  generated  in  it  by  reason  of  the  fact  that 
it  is  "  cutting  "  across  the  lines  of  force. 

If  the  flux  density  produced  by  the  pole  is  B  and  the 
length  of  the  cross  piece  is  /,  and  if  the  cross  piece  is  moving 
to  the  right  at  a  velocity  v,  the  area  of  the  loop  will  be  in- 
creasing at  a  rate 

Iv  square  centimeters  per  second. 
The  flux  through  the  loop  is  thus  increasing  at  a  rate 
Blv  maxwells  per  second. 

Since  the  number  of  turns  in  the  loop  is  unity,  the  voltage 
produced  by  the  motion  of  the  cross  piece  will  accordingly  be 

e  =  Blv  abvolts.  (4) 

Considering  the  fact  that  this  voltage  is  all  generated  in 
the  short  wire,  we  note  that  a  wire  of  length  I  moving  per- 
pendicular to  a  field  of  flux  density  B  and  moving  at  a  ve- 
locity v,  will  have  generated  in  it  a  voltage  equal  to  Blv. 
If  all  quantities  are  expressed  in  units  of  the  c.g.s.  system, 
the  voltage  will  be  in  abvolts.  Converting  to  volts  we  have 

e  =  Blv  10-8  volts.  (5) 

Note  that  it  is  velocity  perpendicular  to  the  field  which 
counts.  The  wire  must  be  actually  cutting  across  the  flux 
lines.  A  conductor  moving  parallel  to  a  field  will  have  no 
voltage  whatever  generated  in  it.  A  conductor  moving  in 
a  direction  oblique  to  the  direction  of  the  field  will  have  in  it 
a  voltage  proportional  to  the  component  of  its  velocity 
perpendicular  to  the  direction  of  the  field. 

The  relation  between  the  direction  of  the  flux,  the  motion 
and  the  voltage  produced  may  be  expressed  conveniently 
by  means  of  Fleming's  right-hand  rule.  Referring  to  Fig. 
164,  this  rule  may  be  expressed  as  follows.  Place  the  right 


332      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

hand  with  the  thumb,  forefinger  and  center  finger  all  at  right 
angles.  Let  the  thumb  point  in  the  direction  in  which  the 
wire  is  moving.  Point  the  forefinger  in  the  direction  of  the 
flux  lines.  The  center  finger  will  then  point  in  the  direction 
in  which  the  voltage  is  produced. 


FIG.  164.  The  thumb  points  in  direction  of  the  motion  of  the  wire,  the 
forefinger  in  the  direction  of  the  flux  and  the  middle  finger  shows  the 
direction  of  the  induced  electromotive  force. 

Applying  this  rule  to  Fig.  163,  in  which  we  found  that  the 
voltage  was  from  back  to  front  of  the  moving  wire,  we  see 
that  the  rule  is  correctly  expressed. 

Prob.  1-10.  When  the  coil  of  one  turn  shown  in  Fig.  161 
occupies  a  position  with  its  plane  at  right  angles  to  the  flux 
lines,  a  total  flux  of  3  X  104  maxwells  links  the  coil.  The  coil 
is  then  turned  at  a  uniform  speed  to  a  position  parallel  with  the 
flux  lines  in  a  time  of  0.005  second.  What  average  voltage  is 
generated  in  the  coil?  What  is  the  instantaneous  voltage  at 
the  beginning  and  at  the  end  of  this  time  interval? 

Prob.  2-10.  A  square  coil  of  10  closely  packed  turns  and  of 
10  square  inches  mean  area  is  rotated  about  a  central  axis 
parallel  to  the  sides  of  the  coil  through  an  angle  of  90°  in  0.0012 
second.  An  average  voltage  of  8  volts  is  generated  in  the  coil 
during  this  interval.  The  coil  is  situated  in  a  uniform  field 


GENERATED   VOLTAGES 


333 


and  is  initially  parallel  to  the  flux  lines.     What  is  the  flux 
density? 

Prob.  3-10.  The  wire  in  Fig.  164  is  moved  through  the 
gap  at  right  angles  to  the  field  at  a  velocity  of  60  feet  a  second 
and  the  voltage  induced  as  it  cuts  the  flux  is  0.85.  The  pole 
faces  are  3  centimeters  wide  and  5  centimeters  long  (the  length 
being  taken  in  the  direction  parallel  with  the  conductor). 
Assuming  no  fringing  of  flux,  what  is  the  air-gap  flux? 


96.  Elementary  Alternators.  Fig.  165  shows  a  cross- 
section  of  the  arrangement  shown  in  Fig.  161.  Suppose 
that  the  coil  is  mounted  on  a  shaft  perpendicular  to  the 
paper  and  caused  to  ^-^ 

revolve  in  this  uniform 
field.  Applying  the 
right-hand  rule,  at  the 
instant  shown,  we  see 
that  in  the  conductor 
on  the  left  there  is  in- 


N 


FIG.  165.    A  cross-section  of  the  ar- 
rangement shown  in  Fig.  161. 


duced  a  voltage  towards 
the  reader,  and  in  the 
other  conductor  a  voltage  away  from  the  reader.  The 
amount  of  this  voltage  is  proportional  to  the  component 
of  the  velocity  in  a  direction  perpendicular  to  the  field. 
This  component  of  the  velocity  is  proportional  to  the 
sine  of  0,  which  is  the  angle  between  the  plane  of  the 
coil  and  a  plane  perpendicular  to  the  flux  lines.  If  the  coil 
is  revolving  at  a  uniform  velocity,  6  is  varying  uniformly  with 
the  time,  and  this  component  of  the  velocity  is  accordingly 
a  sine  function  of  the  time.  The  voltage  produced  is 
then 

e  =  Era&x  sin  ut,  (6) 

where         w  =  the  angular  speed  of  the  coil  in 

radians  per  second, 
j£max  =  maximum  voltage  generated. 

The  maximum  rate  of  cutting  of  flux  and  the  maximum  volt- 
age will  occur  when  the  plane  of  the  coil  is  parallel  to  the 


334      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


flux  lines.     If  v  is  the  peripheral  speed  of  the  coil  we  shall 
have  at  this  instant  a  voltage  generated 

Blv 


108 


volts, 


(7) 


where  I  is  the  length  of  total  conductor  in  the  coil  which  is 
active.     The  length  I  thus  includes  only  the  length  of  the 

wire  which  lies  in  the 
direction  perpendicular 
to  the  paper.  The  end 

sin  at  / 


FIG.  166.    The  form  of  voltage  curve 
induced  in  the  coil  of  Fig,  161  and  165. 


connections  of  the  coil 
which  lie  in  planes  par- 
allel to  the  paper  have 
no  voltage  generated 
in  them  since  they  do 
not  cut  across  the  lines 
of  force. 

This  arrangement  is 
called  an  alternator,  or 
an  alternating-current  generator.  It  produces  a  voltage 
which  varies  with  the  time  in  the  manner  shown  in  Fig. 
166.  The  frequency  of  this 
voltage  is,  in  cycles  per 
second,  the  number  of  revolu- 
tions per  second  of  the  coil 
in  the  arrangement  shown. 

Obviously,  if  such  an  ar- 
rangement as  Fig.  165  were     FIG.  167.    To  decrease  the  reluc- 
USed,  the   reluctance   of   the         tance  of  the  magnetic  path,  the 
air  gap  would  be  large  and       coil  is  wound  on  an  iron  core, 
a     large     magnetizing     coil 

would  be  necessary  in  order  to  force  the  necessary  constant 
working  flux  across  the  long  gap.     To  obviate  this  difficulty 
the  arrangement  used  in  practice  is  more  as  shown  in  Fig. 
167. 
The  revolving  coil  is  situated  in  slots  in  an  iron  armature 


GENERATED   VOLTAGES  335 

which  revolves  as  a  whole.  The  air  gap  between  the  ar- 
mature and  the  pole  pieces  is  small  and  hence  the  reluctance 
of  the  magnetic  circuit  is  kept  down.  The  field  coils,  which 
are  the  magnetizing  coils  for  forcing  the  flux  through  the 
circuit,  will  therefore  require  only  a  small  current  in  order 
to  maintain  the  working  flux.  Placing  the  conductors  in 
slots  in  this  manner  greatly  decreases  the  necessary  air  gap 
and  does  not  affect  the  average  voltage  produced  with  a 
given  speed  and  total  flux.  The  flux  will,  of  course,  be  dis- 
torted and  not  uniform.  The  average  voltage  in  a  conduc- 
tor is,  however,  unchanged,  for  the  total  change  of  linkages 
per  half  revolution  remains  the  same. 

Prob.  4-10.  An  elementary  alternator  such  as  the  one  of 
Fig.  165  has  a  uniform  air-gap  flux  density  of  1000  gausses. 
If  the  coil  revolves  1200  revolutions  per  minute,  what  length 
of  the  conductor  is  necessary  in  the  coil  in  order  to  generate  12 
volts  at  the  maximum  of  the  voltage  wave?  The  wire  used  is 
small  and  when  wound  occupies  a  space  J  inch  square  in  cross- 
section.  The  formed  coil  is  3  inches  square  inside  dimen- 
sions and  3|  outside.  Is  it  allowable  to  use  averages  in  the 
computations? 

97.  The  Direct-Current  Generator.  The  alternator  fur- 
nishes a  potential  which  reverses  in  direction  many  times  a 
second.  For  certain  purposes,  a  generator  is  needed  which 
will  furnish  a  potential  which  does  not  reverse,  that  is,  a 
continuous  potential.  Such  a  direct-current  generator 
employs  a  rectifying  device  called  a  commutator  which 
causes  the  potential  delivered  by  the  machine  to  be  uni- 
directional. 

A  direct-current  generator  is  shown  diagrammatically  in 
Fig.  168.  A  slotted  armature,  built  up  of  steel  laminations, 
is  mounted  on  a  shaft  and  rotated  between  pole  pieces  which 
furnish  a  flux  exactly  as  in  the  alternator.  In  the  figure,  for 
convenience,  are  shown  only  six  armature  slots  with  twelve 
conductors  or  two  conductors  per  slot.  In  actual  machines,  a 
much  larger  number  of  slots  than  this  is  used,  but  a  simple  ar- 
rangement is  shown  in  the  diagram  for  the  sake  of  clearness. 


336      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

These  armature  conductors  are  connected  as  follows. 
The  circuit  passes  up  through  conductor  1,  down  through  2, 
up  through  3,  down  through  4,  up  through  5,  down  through 
6  and  so  on.  The  end  connections  between  the  conductors 
are  shown  by  light  lines,  those  on  the  near  end  of  the  arma- 


FIG.  168.     Diagram  of  a  simple  6-coil  direct-current  generator. 

ture  by  full  lines  and  those  on  the  far  end  by  dotted  lines. 
It  will  be  noted  that  the  circuit  passes  up  through  11,  down 
through  12  and  then  up  through  1,  and  so  on,  so  that  the 
circuit  is  entirely  continuous. 

On  the  same  shaft  with  the  armature,  as  shown  in  the 
side  view,  is  mounted  a  commutator.  It  consists  in  this 
case  of  six  copper  segments,  insulated  from  the  shaft  and  also 
insulated  from  one  another  by  mica  separating  sheets.  The 
cylindrical  surface  of  this  commutator  is  turned  off  to  a 
smooth  bearing  surface.  The  commutator  bars  are  then 
connected  with  the  winding  as  follows:  each  segment  or 
commutator  bar  is  connected  to  one  of  the  end  connections 
passing  between  two  conductors.  Thus  the  segment  labeled 
1-2  is  connected  to  the  end  connections  between  conductors 
1  and  2.  The  positions  on  the  end  connections  where  the 
commutator  bars  are  connected  are  shown  by  dots  in  the 
figure.  The  commutator  is  shown  below  instead  of  in  its 
correct  position  in  front  of  the  armature  for  convenience  in 
drawing. 

Resting  on  the  cylindrical  surface  of  the  commutator  are 
two  brushes  as  shown.  These  are  carbon  blocks  held  in 


GENERATED   VOLTAGES  337 

brush  holders  stationary  on  the  frame  of  the  machine. 
They  rest  upon  the  surface  of  the  commutator  and  make 
electrical  contact  with  it.  To  these  brushes  are  connected 
the  leads  of  the  machine. 

Let  us  now  trace  the  path  of  the  current  through  the  ar- 
mature from  one  brush  to  the  other.  Beginning  at  brush 
A,  we  pass  immediately  to  segment  11-12.  This  is  con- 
nected to  an  end  connection  at  the  point  shown  by  the  dot. 
The  path  then  divides  and  we  have  a  choice  of  two  parallel 
paths  through  the  armature.  Choosing  one  of  these  paths, 
we  pass  down  through  conductor  11,  up  through  10,  down 
through  9,  up  through  8,  down  through  7,  up  through  6, 
then  to  the  segment  marked  5-6  and  the  brush  B,  that  is, 
to  the  other  lead  of  the  machine. 

Consider,  however,  the  direction  in  which  the  voltage  is 
generating  in  the  conductors  due  to  the  revolution  of  the 
armature.  If  the  armature  revolves  in  the  direction  shown 
by  the  arrow,  we  shall  have  a  voltage  generated  down  through 
conductors  11,  9  and  7,  and  up  through  10,  8  and  6.  In  the 
path  from  one  brush  to  the  other  we  thus  have  six  voltages 
generated  which  are  all  so  directed  as  to  force  current  through 
the  circuit  in  the  direction  from  brush  A  to  brush  B.  The 
voltage  between  A  and  B  is  thus  the  sum  of  the  six  voltages 
generated  in  these  separate  conductors. 

If  we  consider  the  other  path  through,  the  armature,  we 
shall  pass  from  brush  A  to  the  segment  11-12,  then,  taking 
the  new  path,  down  through  conductor  12,  up  through  1, 
down  through  2  and  so  on.  Examining  the  voltages  in  this 
path,  we  find  that  they  are  in  the  same  direction  and  of  the 
same  magnitude  as  the  voltages  which  were  encountered 
in  the  other  path.  We  thus  have  between  the  brushes  A 
and  B  two  parallel  paths  through  the  armature  in  each  one 
of  which  an  equal  voltage  is  generated  in  a  direction  from 
brush  A  to  brush  B.  Brush  B  will  hence  be  positive  and 
brush  A  negative. 

Let  us  now  consider  what  happens  to  the  connections  as 


338      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

the  armature  revolves.  Consider  the  instant  when  the 
armature  has  turned  through  one-sixth  of  a  revolution  be- 
yond the  position  shown  in  the  figure.  Brush  A  will  then 
rest  on  segment  9-10  and  brush  B  on  segment  3-4.  That  is, 
the  connections  to  the  armature  are  changed  by  the  com- 
mutator. In  passing  from  brush  A  to  brush  B,  however, 
we  find  upon  examination  that  we  go  by  exactly  the  same 
paths  as  before,  if  we  consider  simply  the  positions  of  the 
conductors.  The  conductors  have  all  moved  around  one 
position,  but  new  conductors  have  taken  their  places. 
In  going  from  A  to  B,  we  again,  however,  pass  down  through 
a  conductor  under  a  south  pole,  and  so  on.  This  is  exactly 
what  we  did  before,  and  so  the  voltage  generated  is  exactly 
the  same  as  before  and  in  the  same  direction.  After  an- 
other one-sixth  revolution  the  connections  will  again  change, 
but  we  shall  have  as  before  exactly  the  same  voltage  between 
brushes.  The  potential  developed  by  the  machine  ac- 
cordingly does  not  change  in  direction  as  the  armature 
revolves;  that  is,  it  is  a  direct-current  generator. 

In  the  machine  shown  in  the  figure,  with  only  six  slots, 
the  voltage  will  vary  somewhat  as  the  armature  revolves, 

due    to    the    variation    in 
position    of    the   armature 
conductors  during  the  time 
that   the   brush  is  passing 
over     a     single     segment. 
The   voltage    delivered  by 
the  machine  will  hence  have 
somewhat   the   appearance 
FIG.  169.     The  curve  of  voltage       shown  in  Fig.  169.     It  will 
delivered  to  the  brushes  of  the       be   unidirectional,   that   is, 
generator  in  Fig.  168.  it  will  not  reverse,   but  it 

will  have  a  very  consider- 
able ripple.  In  an  actual  machine  made  with  perhaps  100 
slots,  this  ripple  will  be  almost  entirely  removed,  and  the 
voltage  will  appear  as  shown  in  Fig.  170,  for  the  variation 


GENERATED  VOLTAGES 


339 


in  position  of  the  conductors  under  the  poles  during  the 
time  that  the  brush  is  passing  over  a  single  segment  is  al- 
most entirely  negligible. 

The  machine  shown  in  diagram  in  Fig.  168  is  a  two-pole 
machine.  A  much  more  common  form  in  practice  is  a 
multipolar  machine,  having 
four  or  six  poles  or  even 
more.  The  simplest  form 
of  drum  winding  is  also 
shown  on  the  armature  in 
the  diagram.  There  are 
many  forms  of  armature 
winding,  all  of  which  are 
designed  to  give  the  same 
result  as  the  winding  shown 


FIG.  170.  The  terminal  voltage 
of  a  direct-current  generator 
having  a  large  number  of  com- 
mutator segments. 


in    the    figure,   that    is,   to 

connect   between  brushes  a 

series  of  conductors  in  such  a  manner  that  the  voltages 

'always   add.     Fig.  171   shows  the  armature  of  a   small 


FIG.  171.     The  armature  of  a  direct-current  generator.    General 
Electric  Co. 

commercial  generator,  the  poles  and  frame   of  which  are 
shown  in  Fig.  172. 

The  construction,  winding  and  performance  of  direct- 
current  machinery  are  too  large  subjects  to  be  taken  up  here. 
We  can  consider  at  this  time  only  the  basic  principles  under- 
lying all  such  machinery. 


340      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  5-10.  On  the  armature  of  a  certain  8-pole  direct- 
current  generator,  there  are  146  active  conductors  in  series 
between  the  brushes.  Each  pole  has  1.2  X  106  lines  of  flux. 
The  speed  is  1800  revolutions  per  minute.  What  is  the  average 
generated  voltage  between  brushes? 


FIG.  172.    Partly  assembled  frame  of  a  four-pole  direct-current 
generator.    General  Electric  Co. 

Prob.  6-10.  It  is  desired  to  design  a  generator  which  will 
generate  an  average  electromotive  force  of  230  volts.  An 
armature  is  to  be  used  which  can  give  180  active  conductors 
between  brushes.  How  many  poles  must  be  used,  each  with 
3  X  106  lines  of  flux,  in  order  to  allow  the  machine  to  run  at 
about  480  revolutions  per  minute? 

Prob.  7-10.  At  a  flux  density  at  the  pole  surface  of  65,000 
lines  per  square  inch,  how  large  must  the  poles  of  a  6-pole 
generator  be  if  it  is  to  generate  an  average  electromotive  force 
of  115  volts?  There  are  208  conductors  between  brushes  and 
the  speed  is  900  revolutions  per  minute. 

98.  Conductor  in  a  Moving  Field.  In  order  to  generate 
a  voltage  we  must  move  a  conductor  relative  to  a  magnetic 
field.  The  voltage  then  produced  will  be  proportional  to  the 
length  of  the  conductor,  the  strength  of  the  field  and  the 
component  of  the  velocity  of  the  conductor  perpendicular 
to  the  field.  It  does  not  matter,  however,  whether  we  move 


GENERATED   VOLTAGES 


341 


the  conductor  and  keep  the  magnetic  .field  stationary,  or 
whether  we  keep  the  conductor  stationary  and  move  the 
magnetic  field.  It  is  only  relative  motion  that  counts.  The 
same  voltage  will  be  generated  in  either  case  provided  the 
relative  velocity  is  the  same. 

This  principle  is  made  use  of  in  the  revolving-field  alter- 
nator. In  this  machine,  which  is  the  usual  form  of  prac- 
tical commercial  alternator,  the  armature  windings  are 
stationary  and  the  field 
structure  revolves. 

A  simple  form  of  re- 
volving-field alternator 
is  shown  diagrammatic- 
ally  in  Fig.  173.  A  coil, 
A  A,  is  placed  in  slots 
in  a  stationary  iron  yoke 
called  a  stator.  This 
stator  is  mounted  rigidly 
on  the  frame.  Inside  of 

the  stator  revolves  a  large 

,  11    i  ji         FIG.  173.      The  diagram  of  a  three- 

electromagnet  called  the        phage    alternator.8   The    armature 

revolving  field.  In  the  stands  stai)  while  the  field  revolves, 
diagram  a  two-pole  field 

is  shown.  This  magnet  receives  its  exciting  current,  that 
is,  its  magnetizing  current,  by  means  'of  two  slip-rings 
which  are  simply  insulated  metal  rings  mounted  on  the 
shaft,  to  which  current  is  conducted  by  brushes  sliding  on 
their  surface. 

The  field  magnet  as  it  revolves  carries  its  flux  with  it. 
This  flux  is  thus  caused  to  move  across  the  coil  A .  A  voltage 
will  then  be  set  up  in  this  coil,  and  its  magnitude  may  be 
computed  from  the  voltage  formula  above.  This  voltage 
will  be  alternating,  for  since  the  direction  of  the  flux  passing 
the  conductor  changes  as  the  second  pole  comes  under  the 
conductor,  the  direction  of  the  voltage  generated  will  also 
change. 


342      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

With  a  concentrated  coil  mounted  in  a  single  slot  as 
shown,  the  voltage  generated  at  any  instant  will  be  propor- 
tional to  the  flux  density  at  that  time  passing  the  coil.  The 
voltage  in  the  coil  will  accordingly  vary  in  exactly  the  same 
manner  that  the  flux  density  varies  across  the  face  of  the 
field  pole.  The  voltage  generated  may  hence  have  some- 
what the  form  shown  in  Fig.  174.  If  the  air  gap  is  made 


\J 


FIG.  174. 


Curve  of  voltage  generated  in  one  coil  of  the  armature 
of  Fig.  173. 


larger  near  the  pole  tip  than  at  the  center,  then  the  shape  of 
the  curve  of  flux  density  across  the  pole  face  may  be  made 
nearly  sinusoidal.  In  this  case  the  delivered  voltage  will 
also  be  nearly  sinusoidal. 

Suppose  that  we  make  more  slots  in  the  stator  and  place 
coils  as  shown  at  BB  and  CC.  These  coils  will  also  have 
voltages  generated  in  them,  and  voltages  which  will  alter- 
nate in  exactly  the  same  manner  as  does  the  voltage  in  coil 
A.  The  voltage  in  B  will,  however,  come  to  its  maximum 
value  a  certain  definite  time  later  than  the  voltage  in  A. 
Similarly  coil  C  will  attain  its  maximum  at  a  still  later  time. 
Accordingly,  if  we  assume  that  the  flux  distribution  across 
the  pole  face  is  sinusoidal,  the  voltages  delivered  by  these 
three  coils  may  be  plotted  on  the  same  diagram  as  shown 
in  Fig.  175  and  will  have  the  relations  shown. 


GENERATED   VOLTAGES 


343 


Such  a  machine  is  called  a  three-phase  generator.  It  is 
really  three  generators  combined  into  one.  Each  one  de- 
livers an  alternating  potential,  but  they  are  not  in  step. 
They  come  to  a  maximum  successively,  pass  through  zero 
successively  and  so  on.  Most  commercial  alternators  are 
of  the  three-phase  type. 


FIG.  175.  Curves  of  the  voltages  generated  in  the  three  coils  of  a 
generator  like  that  shown  in  Fig.  173.  The  flux  distribution  has  been 
changed  from  that  of  Fig.  174  so  that  sine-wave  forms  are  generated. 

In  the  figure  we  have  shown  each  coil  side  entirely  situated 
in  a  single  slot.  Such  an  arrangement  is  called  a  concen- 
trated coil  winding.  It  is,  however,  more  usual  to  dis- 
tribute the  coils  through  several  adjacent  slots.  Such  a 
distributed  coil  winding  will  give  much  better  wave  form 
than  the  concentrated  form  shown. 

The  two-pole  arrangement  shown  in  Fig.  173  is  almost 
never  used,  except  in  high-speed  turbo-alternators.  Low- 
speed  alternators  are  always  multipolar  in  type.  Engine- 
driven  alternators  may  contain  a  large  number  of  poles,  as 
high  as  eighty.  Fig.  176  shows  the  rotor  of  a  vertical  type 
of  water- wheel-driven  alternator  having  forty-eight  poles. 
Fig.  177  shows  the  stator  or  armature  of  the  same  machine. 

The  revolving-field  type  of  alternator  is  almost  always 
used  today,  and  principally  for  the  following  reasons.  It 
will  be  noted  that  the  coils  in  which  the  voltage  is  generated 
may  be  connected  directly  to  the  external  circuit,  since  they 
are  stationary.  No  collecting-rings  are  necessary  to  com- 


344      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

plete  the  circuit.     This  is  advantageous  in  machines  where 
a  high  voltage  is  generated.     Large  alternators  today  usu- 


FIG.    176.     The  rotating  field  of  a  large  water-wheel-driven  alternator. 

ally  generate  directly  voltages  of  from  2300  to  13,000  volts. 
It  is  not  at  all  convenient  to  handle  such  voltages  on  col- 
lecting-rings. In  addition  the  large  current  delivered  by 


GENERATED   VOLTAGES 


345 


modern  high-output  al- 
ternators will  not  have 
to  be  carried  by  collect- 
ing-rings. Alternators 
having  outputs  as  high 
as  45,000  kilovolt- 
amperes  on  a  single 
machine  are  built,  and 
this  means  very  large 
currents  indeed.  The 
revolving-field  type  also 
has  certain  advantages 
as  regards  strength  of 
construction. 

Prob.  8-10.  The  re- 
volving-field alternator 
of  Fig.  173  has  8  con- 
ductors in  each  of  the 
slots  "  C."  The  con- 
ductors are  connected  in 
series  so  that  the  gen- 
erated voltages  are 
added.  The  effective 
length  of  the  slot  is  8.5 
inches  and  the  speed  of 
the  rotor  is  1200  revolu- 
tions per  minute.  The 
radius  from  the  center 
of  the  shaft  to  the  center 
of  the  bundle  of  con- 
ductors in  the  slot  is  6.1 
inches.  What  maximum 
flux  density  is  necessary 
in  the  gap  to  generate  25 
volts  maximum? 

99.  The   Homopolar 

Generator.      It    is    en-   FIQ.  177.    The  stationary  armature  for 
tirely   possible  to    con-       the  rotating  field  shown  in  Fig.  176. 


346      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

struct  a  direct-current  generator  without  using  a  com- 
mutator. Such  machines  may  be  made  by  causing  a 
conductor  to  move  continuously  through  a  field  of  constant 
intensity  and  always  in  the  same  direction.  They  are  called 
homopolar  generators.  They  are  not  much  used  except 
where  very  high  currents  are  desired  at  low  voltages,  and  are 
manufactured  almost  exclusively  for  this  purpose. 

Such  a  homopolar  generator  is  shown  diagrammatically  in 
Fig.  178a  and  b.    The  machine  is  shown  in  section.    It  is  made 


FIG.  178a.     Diagram  of  a  homopolar  generator.     No  commutator 
is  needed. 

of  an  iron  piece  having  a  cylindrical  air  gap.  Field  coils  shown 
in  section  as  FI  and  F2  are  stationary  and  carry  a  current 
which  forces  a  flux  through  the  stationary  iron  structure  as 
shown  by  arrows.  This  flux  thus  passes  out  radially  across 
the  air  gap. 

Metal  rings  shown  at  Si  and  S2  are  mounted  on  insulating 
spiders  so  that  they  may  be  revolved.  Between  these  rings 
is  connected  a  heavy  conductor  which  thus  revolves  with 
them  around  through  the  cylindrical  air  gap.  The  current 
is  collected  by  means  of  brushes  resting  on  the  slip-rings. 
These  brushes  are  held  in  stationary  brush  holders.  Ac- 


GENERATED   VOLTAGES 


347 


cess  may  be  had  to  the  brushes  by  means  of  hand-holes 
made  in  the  iron  structure.  It  will  be  noted  that  the  con- 
ductor is  always  passing  through  a  field  of  constant  intensity, 
and  always  perpendicular  to  this  field.  It  accordingly  has 
a  constant  potential  generated  in  it. 

By  using  a  large  magnetic  flux  and  a  high  speed,  a  voltage 
as  high  as  40  volts  may  be  obtained  from  a  single  conductor. 
Several  conductors  are  generally  used,  however,  placed  at 
intervals  around  the  air  gap.  Each  one  is  connected  to 
two  entirely  separate  collect- 
ing-rings, so  that  there  are 
twice  as  many  collecting- 
rings  as  there  are  conductors. 
By  external  connections  with 
the  brushes,  these  conductors 
are  then  placed  in  series  as  far 
as  the  external  circuit  is  con- 
cerned. Their  voltages  thus 
add  and  the  voltage  delivered 
by  the  machine  is  the  sum  of 
the  voltages  of  the  separate 
conductors. 

The   machine    requires    no 

commutator  and  delivers  a  constant  direct  voltage.  It 
has,  however,  certain  obvious  structural  *  difficulties  which 
have  prevented  its  adoption  for  any  purposes  except  those 
noted  above. 

Prob.  9-10.  The  effective  conductor  length  in  Fig.  178a  and  b 
is  9.3  inches.  The  conductors  are  at  a  radius  of  4  inches  from  the 
center  line  of  the  shaft.  Eight  conductors  are  used  connected 
in  series  by  means  of  slip-rings.  At  a  speed  of  3600  revolutions 
per  minute  and  a  total  air-gap  flux  of  8  X  106  maxwells,  what 
voltage  is  generated  in  the  machine? 

100.  Eddy  Currents.  Whenever  the  amount  of  flux  link- 
ing a  circuit  changes,  there  is  a  voltage  set  up  in  the  circuit, 
and  if  the  circuit  is  closed,  a  current  flows.  All  materials 


FIG.    1786.      Section  A- A  of 
the  view  shown  in  Fig.  178a. 


348      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

have  resistance,  at  least  at  ordinary  temperatures.  Ac- 
cordingly whenever  a  current  flows  in  a  conductor  there  is 
a  loss  which  appears  as  a  heating  of  the  conductor.  Any 
variation  of  flux  therefore  always  entails  losses  in  circuits 
which  the  flux  links. 

It  is  not  necessary  that  the  circuit  be  a  wire  and  that  the 
flux  pass  entirely  through  it.  If  a  solid  block  of  metal  is 
traversed  by  a  varying  flux,  metallic  circuits  in  the  block 
itself  which  are  linked  by  the  flux  will  carry  current.  If  a 
magnetic  circuit  is  made  up  of  iron  and  if  the  flux  in  the 
circuit  is  variable,  there  will  be  current  set  up  by  induction 
in  the  iron  of  the  circuit  itself.  All  of  these  stray  currents 
entail  losses,  and  such  losses  are  known  as  eddy-current 
losses. 

Electrical  machinery  usually  involves  varying  fluxes. 
In  alternating-current  work,  the  transformer  depends  for 
its  action  upon  the  variation  of  flux  in  a  core.  In  direct- 
current  machinery  as  well  as  alternating-current  machin- 
ery, we  have  an  armature  built  of  iron  which  is  either  re- 
volved in  a  permanent  field,  or  through  which  a  revolving 
field  passes.  In  either  case  there  will  be  losses  in  the  iron 
due  to  hysteresis  and  also  losses  due  to  eddy  currents. 
The  toothed-core  armature  of  a  direct-current  machine 
causes  tufts  of  flux  to  sweep  across  the  pole  faces,  thereby 
setting  up  eddy  currents  in  the  iron  of  the  poles.  In  a 
dynamo,  in  fact,  there  will  be  eddy-current  losses  in  the 
conductors  themselves  if  they  are  of  large  size.  In  addition 
to  the  main  voltage  which  is  set  up  in  the  conductors,  there 
will  be  locally  generated  voltages  due  to  variation  in  the 
density  of  flux  which  will  cause  local  current  to  circulate  in 
the  material  of  the  conductors  without  passing  through  the 
external  circuit. 

Eddy  currents  always  tend  to  flow  in  planes  perpendicular 
to  the  flux,  since  they  are  set  up  by  the  variation  of  this  flux 
through  a  circuit.  If  the  material  of  which  the  magnetic 
circuit  is  made  is  subdivided  into  thin  sheets,  that  is,  if  the 


GENERATED   VOLTAGES  349 

material  is  laminated,  and  these  sheets  are  placed  parallel 
to  the  flux  and  insulated  from  each  other,  the  eddy-current 
loss  will  be  reduced.  The  subdividing  of  the  iron  into  in- 
sulated sheets  in  this  manner  breaks  up  the  paths  in  which 
eddy  currents  flow,  and  greatly  reduces  the  magnitude  of  the 
eddy  currents. 

If  a  direct-current  generator  were  made  with  an  armature 
constructed  of  a  solid  piece  of  soft  iron,  the  eddy-current  loss 
would  be  enormous.  In  fact,  some  early  generators  thus 
constructed  required  several  horse  power  to  drive  them  while 
the  useful  output  of  the  machine  was  only  a  small  fraction 
of  a  kilowatt.  Almost  all  of  the  input  of  the  machine  was 
lost  in  the  stray  currents  produced  in  the  armature  by  the 
variation  of  the  flux  density  at  various  points  as  the  arma- 
ture revolved.  In  order  to  avoid  this  large  loss,  armatures 
are  made  up  of  sheet-steel  laminations,  usually  between  0.014 
and  0.030  inch  thick,  assembled  on  a  shaft,  so  that  the  ar- 
mature is  subdivided  into  sheets  which  are  parallel  to  the 
working  flux  of  the  machine.  Eddy  currents  which  tend  to 
flow  perpendicular  to  this  flux  are  thus  reduced  to  a  small 
value  by  the  high  resistance  between  the  sheets. 

In  ordinary  direct-current  generators,  no  special  pre- 
caution is  taken  to  insulate  these  sheets  from  one  another, 
for  the  scale  which  is  naturally  on  the  sheet  steel  after  it  is 
rolled  prevents  good  electrical  contact  from  taking  place. 
In  constructing  transformers,  the  sheets  are  often  given  a 
coat  of  shellac  or  varnish  in  order  to  insulate  them.  Care 
must  be  taken,  of  course,  that  in  bolting  the  sheets  together 
they  are  not  electrically  connected  to  one  another.  Also 
filing  the  edges  of  a  core,  or  the  slot  of  an  armature,  will 
greatly  increase  eddy-current  loss  by  bringing  the  successive 
laminations  into  electrical  contact.  In  high-frequency 
machines  or  transformers,  great  care  indeed  must  be 
taken  to  keep  down  the  eddy-current  losses.  In  generators 
delivering  frequencies  of  50,000  cycles  per  second,  iron 
of  a  thickness  of  approximately  0.0015  inch  is  used,  and 


350      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


is  carefully  japanned  in  order  to  prevent  contact  between 
the  sheets. 

The  following  analysis  will  show  the  magnitude  of  the 
eddy-current  loss  to  be  expected,  as  well  as  the  effect  of 
subdividing  the  material  into  thin  sheets,  the  effect  of  re- 
sistivity and  so  on. 

Let  us  consider  first  the  eddy-current  loss  in  a  ring  which 
completely  surrounds  a  magnetic  circuit. 
Such  an  arrangement  is  shown  in  Fig. 
179  and  a  long,  thin  copper  ring  is  shown 
because  of  its  bearing  on  the  losses  in 
laminations  to  be  considered  below.     The 
iron  piece  which  threads  this  ring  we  will 
consider  as  carrying  a  flux  <£.     This  flux 
^3[/'  copper      *s  varvmS>  and  due  to  its  variation  there 
s'  is  a  voltage  generated  which  causes  cur- 

rent to  flow  around  the  short-circuited 
copper  ring.  The  amount  of  loss,  that 
is,  the  PR  in  the  ring,  caused  by  this 
current  is  to  be  computed. 

We  will  consider  the  ring  to  be  long 
and  narrow,  so  that  the  resistance  of  its 
ends  can  be  neglected  in  comparison  with  the  resistance  of 
the  sides.  The  resistance  then  is 


FIG.  179.  A  thin 
copper  strip  sur- 
rounding an  iron 
bar  in  which  the 
magnetic  flux  is 
changing. 


R  =  p  —£  ohms , 
ab 


(8) 


where  p  is  the  resistivity  of  the  material  of  which  the  ring  is 
constructed.  If  I,  a  and  6  are  in  centimeters,  then  p  must 
be  in  ohms  per  centimeter  cube  of  the  material. 

When  the  flux  is  changing,  there  will  be  a  voltage  gener- 
ated of  value 


dt 


10-8  volts, 


(9) 


GENERATED   VOLTAGES 


351 


and,  by  Ohm's  law,  this  will  cause  a  current  to  flow  as  given 
by 


=—z-1 10~8  amperes. 

p- 
ab 


(10) 


The  watts  loss  at  any  instant  due  to  this  current  is  the 
product  of  the  square  of  the  current  and  the  resistance,  that 
is. 


p  = 


/d*y 
\dt) 


"a-fc 


10-16  watts. 


(11) 


The  electrical  machinery  in  which  we  are  usually  interested 
has  fluxes  that  vary  sinusoidally  with  the  time.    That  is, 


FIG.  180.     Curve  showing  how  the  flux  in  the  iron  bar  of 
Fig.  179  varies. 

we  are  interested  in  a  flux  which  varies  in  such  a  manner 
that  its  value  at  any  instant  is  given  by  the  equation 

</>  =  </>max  sin  2irft  maxwells.  (12) 

This  manner  of  variation  is  shown  in  Fig.  180.  /  is  the 
frequency  of  the  flux;  that  is,  it  is  the  number  of  times  the 
flux  changes  through  a  complete  cycle  per  second.  Ex- 


352     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

amining  this  figure,  we  see  that  when  t  =  0,  the  flux  is 
zero  and  is  increasing.  When  t  =  I//,  the  flux  is  again 
zero  and  is  again  increasing.  The  quantity  2ir  ft  is  always 
to  be  taken  in  radians,  and  the  sine  of  2ir  radians  is  zero. 
T  or  I//  is  therefore  called  the  period  of  the  alternating 
flux,  since  it  is  the  time  in  which  the  flux  makes  a  complete 
change. 

When  a  flux  is  varying  in  this  manner,  its  rate  of  change 
may  be  found  for  any  instant  by  differentiating  the  above 
expression,  giving 

-^  =  2x/0max  cos  2*  ft  maxwells  per  second.      (13) 

We  may  insert  this  value  for  the  rate  of  change  of  flux 
in  the  expression  for  the  i2R  loss  at  any  instant,  obtaining 

X  1Q~16  C0g2  2Trft  watts         (14) 


21 
pai> 

=  Pmax  COS2  2w  ft  Watts, 

where  Pmax  is  an  abbreviation  introduced  for  convenience  to 
denote  the  maximum  value  to  which  the  instantaneous  loss 

attains.  The  loss  thus 
varies  from  instant  to 
instant  in  accordance 
with  a  cosine-squared 
___^ term.  The  manner  in 

\l  \  which  the  loss  varies  is 

/  \     /       obtained  by  plotting  this 

expression  as  is  done  in 


p 


7\    A 


FIG.  181.    Curve  of  power  loss  due  to     Flf '  181' 
eddy     currents    when    flux    varies          ln.  Order   to    nnd   the 
sinusoidally.  loss  in  the  ring,  we  are, 

of  course,   interested  in 

the  average  value  of  the  i2R  loss,  that  is,  in  the  average 
value  of  the  above  expression.     We  can  obtain  this  average 


GENERATED  VOLTAGES  353 

value  by  integrating  the  expression  over  a  complete  period 
and  dividing  by  the  length  of  the  period.     That  is, 

(15) 

Inserting  into  this  expression  the  value  of  p,  remembering 
that  T  =  y> 


r- 
=  //' 

Jo 


Dividing  /  by  2irf,  and  multiplying  the  factor  under  the 
integration  sign  by  2-jrf  in  order  not  to  change  the  value  of 
the  expression,  we  obtain 


x   ft 

JQ 


COS2  (2wft)    (27T/)  dt.  (16) 


To  evaluate  this  expression,  we  may  make  use  of  the  known 
integration 

„     ,        x  .  sin  2x 
cos2z  dx  =  ^  H j—  •  (17) 

In  our  expression,  x  is  replaced  by  2ir}t.     Inserting  this 
value,  we  obtain 


..o'  (18) 

Upon  substituting  the  limits,  the  second  term  of  this  ex- 
pression becomes  zero  for  each  limit.  The  entire  expression 
thus  reduces  to 

p      -     P 

•fay  —  Q  *  max  * 

This  means  that  the  average  height  of  the  curve  in  Fig. 
181  is  just  one-half  of  its  maximum  height.  The  average 
i2R  loss  in  this  ring  is  half  of  the  maximum  value  to  which 
it  attains  during  the  cycle. 


354      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 
Insert  the  value  of  PmaX  in  this  expression  and  it  becomes 

Pav  =  f2  0max2  10~16  Watts.  (20) 


This  expression  gives  the  average  watts  loss  in  the  ring 
due  to  the  variation  of  flux  through  it.  Several  interesting 
points  are  to  be  noted  immediately.  The  watts  loss  varies 
as  the  square  of  the  frequency.  Thus  if  the  flux  is  varying 
at  the  rate  of  fifty  cycles  per  second,  there  will  be  four  times 
as  much  loss  as  if  the  same  flux  were  varying  at  the  rate  of 
twenty-five  cycles  per  second.  The  loss  also  varies  as  the 
square  of  the  maximum  flux.  If  the  flux  density  is  doubled, 
the  loss  in  the  ring  will  be  made  four  times  as  great.  The 
loss  also  varies  inversely  as  the  resistivity  of  the  material. 
The  lower  the  resistivity,  the  higher  the  loss  will  be.  This 
may  at  first  sight  seem  peculiar,  but  when  it  is  remembered 
that  decreasing  the  resistance  increases  the  current,  and 
that  the  loss  goes  up  as  the  square  of  the  current,  the  reason 
will  be  clear.  There  will  thus  be  much  more  loss  in  a  short- 
circuited  ring  of  copper  around  a  magnetic  circuit  than, 
for  instance,  in  a  short-circuited  ring  of  brass.  We  must 
remember,  of  course,  that  for  the  maximum  flux  density  we 
are  to  take  the  value  of  the  flux  density  which  actually 
passes  through  the  magnetic  circuit.  The  amount  of  this 
flux  density  will  be  affected  not  only  by  the  magnetomotive 
force  acting  on  the  circuit  but  also  by  the  current  flowing 
in  the  short-circuited  ring  itself.  The  effect  of  this  current 
is  to  tend  to  cause  less  flux  to  pass  through.  A  heavy,  short- 
circuited  ring  of  copper  will  therefore  often  have  less  loss 
for  the  reason  that  the  heavy  current  flowing  will  reduce  the 
net  amount  of  flux  passing  through  the  circuit.  If  the  same 
flux  actually  passes,  however,  the  copper  ring  will  have  a 
higher  loss  than  a  ring  of  greater  resistivity. 

Rings  of  copper  forming  a  short-circuited  path  are  used 
occasionally  in  alternating-current  apparatus.  For  in- 
stance, in  alternating-current  contactors  such  rings  are  used 


GENERATED  VOLTAGES 


355 


to  keep  the  contact  from  chattering.  We  are  more  inter- 
ested, however,  in  an  eddy-current  loss  which  occurs  when 
the  conducting  path  does  not  surround  the  magnetic  circuit, 
but  rather  forms  a  part  of  the  magnetic  circuit  itself;  that 
is,  we  are  more  interested  in  the  losses  which  occur  in  the 
sheet-steel  laminations  of  which  the  magnetic  circuit  is 
built. 

It  has  been  found  experimentally  that  the  same  loss 
occurs  when  a  lamination  is  revolved  in  a  constant  field  of 
a  certain  flux  density  as  will  occur  if  an  alternating  flux 
of  this  maximum  value  is  passed  through  the  sheet.  We 
will  accordingly  analyse  only  the  latter  case.  Our  con- 
clusions will,  however,  apply 
not  only  to  alternating-current 
transformers,  but  also  to  the 
armatures  of  dynamos. 

A  piece  of  iron  lamination 
is  shown  in  Fig.  182.  The  flux 
passes  in  the  direction  of  the 
arrow  and  parallel  to  the  plane 
of  the  sheet,  and  is,  we  will 
consider,  of  maximum  flux  den- 
sity Bmax.  Eddy  currents  will  B 
tend  to  flow  around  through  the 
sheet  in  paths  similar  to  the 
one  drawn  on  the  face  of  the 
sheet.  If  the  thickness  of  the 
lamination  is  small  compared  to 

its  width,  these  paths  may  be  considered  as  parallel 
to  the  surface  of  the  sheet  without  serious  error.  More- 
over, in  computing  the  resistances,  we  can  with  suffi- 
cient precision  neglect  the  resistance  of  the  small  end  por- 
tion of  the  path.  Our  analysis  is  hence  only  approximate. 
As  we  have  noted  above,  however,  various  practical  matters, 
such  as  the  manner  in  which  the  sheets  are  bolted,  greatly 
affect  the  eddy-current  loss.  Since  these  accidental  vari- 


FIG.  182.  Diagram  of  eddy- 
current  circuits  in  a  thin 
sheet  of  iron. 


356      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

ations  may  amount  to  fifty  percent  or  even  more,  it  is 
useless  for  us  to  make  a  too  careful  theoretical  analysis. 
Our  assumptions  then  are  justified,  and  we  can  expect  to 
obtain  as  a  result  of  our  analysis  an  indication  of  the  way  in 
which  the  loss  will  vary,  rather  than  an  accurate  formula  for 
computing  its  value. 

Consider  the  path  at  a  distance  x  from  the  center  line  of 
the  sheet.  Let  the  thickness  of  the  path  be  dx.  There  will 
be  a  voltage  generated  around  this  path  due  to  the  amount  of 
flux 

$x  =  2xlB  maxwells,  (21) 

which  links  through  the  path. 

Conditions  for  this  path  are  now  exactly  the  same  as  for 
the  ring  shown  in  Fig.  179,  if  we  consider  simply  the  portion 
of  the  flux  which  passes  inside  of  the  chosen  path.  The  loss 
in  this  portion  of  the  sheet  will  therefore  be  given  by  formula 
(20)  which  we  have  already  derived,  and  the  fraction  of  the 
total  loss  which  occurs  in  this  path  is 


72  (2xlBm&xy  10-1'  (22) 

/yt/ 

or 


<2Pav   =  ^^  #max2  10~16  X2dx  Watts.  (23) 

P 

The  total  loss  in  the  laminations  will  be  obtained  by 
adding  the  losses  for  all  the  possible  paths  similar  to  the 
above.  In  order  to  do  this,  we  integrate  the  partial  loss 
where  x  varies  from  zero  to  half  the  width  of  the  sheet, 
which  gives 

(24) 

0 

6p~" 

However,  the  volume  of  the  laminations  is  given  by 

v  =  blh  cubic  centimeters.  (25) 


GENERATED  VOLTAGES  357 

The  loss  per  cubic  centimeter  of  the  material  is  obtained 
by  dividing  the  total  loss  by  the  volume  and  hence  be- 
comes 

2 

Pav  =  J-  /*2/2£max2 10~16  watts  per  cubic  centimeter.     (26) 
bp 

This  formula  gives  the  watts  loss  per  cubic  centimeter 
of  a  transformer  core  or  a  dynamo  armature  when 

h  =  the  thickness  of  the  laminations, 
p  =  the  resistivity  of  the  laminations, 
/  =  the  frequency  of  the  flux, 
Bmax  =  the  maximum  flux  density. 

All  the  quantities  in  this  formula  are  to  be  expressed,  of 
course,  in  c.g.s.  units. 

In  practical  cases  this  formula  will  usually  give  results 
which  are  too  low,  owing  to  imperfect  insulation  between 
the  sheets.  It  is  often  written  in  the  form 

pav  =  €  (/i/Bmax)2  10~16  watts  per  cubic  centimeter    (27) 

and  the  constant  e  is  obtained  from  experience.  If  all 
quantities  are  in  c.g.s.  units,  a  reasonable  value  of  the 
coefficient  e  for  good  transformer  steel  is  about  65,000. 
Values  of  this  coefficient  will  be  found  in  all  electrical  hand- 
books. 

Note  that  eddy-current  loss  in  the  laminated  sheets 
varies  as  the  square  of  the  thickness  of  the  sheets.  That  is, 
if  we  have  a  transformer  built  of  0.024  sheet  steel,  and  we 
construct  it  with  the  same  volume  of  iron  but  of  0.012  sheet 
steel  of  the  same  grade,  then  the  eddy  current  will  be  one 
quarter  of  what  it  was  before.  If  the  sheets  are  made 
too  thin,  a  great  amount  of  expense  is  naturally  involved  in 
handling  them,  and  the  space  factor  is  poor.  By  the  space 
factor  is  meant  the  volume  of  actual  iron  divided  by  the 
total  volume  of  the  core.  Increasing  the  insulation  between 
laminations  also  reduces  the  space  factor.  The  choice  of  a 


358      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


proper  thickness  of  steel  for  constructing  a  core  is  usually  a 
balance  between  the  losses  to  be  expected  and  the  cost  of 
construction. 

The  loss,  it  will  be  seen,  varies  as  before  with  the  square  of 

the  frequency  and  with  the 
square  of  the  maximum  flux 
density. 

The  loss  also  varies  in- 
versely as  the  resistivity  of 
the  material.  Good  silicon 
steel  containing  several  per- 
cent of  silicon  will  have  a 
resistivity  about  three  times 
as  great  as  ordinary  trans- 
former steel.  This  is  one 
of  the  principal  reasons 
for  using  silicon  steel  in  the 
construction  of  transformer 
cores  and  similar  apparatus. 
The  eddy  currents  set  up 
in  the  aluminum  disk  of  the 
meter  shown  in  Fig.  182a  are 
used  to  regulate  the  speed 
at  which  the  disk  rotates. 


FIG.  182a.  A  watt-hour  meter. 
The  eddy  currents  set  up  in  the 
disk  as  it  revolves  between  the 
poles  of  permanent  magnets 
regulate  the  speed  of  the 
meter.  The  General  Electric  Co. 


Prob.  10-10.  If  in  Fig.  179  the  bar  of  iron  carries  a  sinu- 
soidally  varying  flux  $max  =  5600  lines,  /  =  50  cycles  per  second, 
what  will  be  the  power  loss  in  a  copper  ring  where  I  =  3  inches, 
o=|  inch  and  b  =  f  inch,  at  a  temperature  of  20°  C.?  If 
the  temperature  of  the  ring  rises  to  150°  C.,  what  will  be  the 
loss  at  this  new  temperature?  How  many  watts  per  square 
inch  of  external  surface  of  the  ring  in  each  case? 

Prob.  11-10.  A  transformer  core  has  a  maximum  flux 
density  of  62,000  lines  per  square  inch  and  a  lamination  thick- 
ness of  0.010  inch  (No.  32  gauge).  With  the  same  net  volume 
of  iron,  if  the  laminations  were  made  0.016  inch  thick  (No.  28 
gauge),  what  change  would  have  to  be  made  in  the  flux  density 
in  order  that  the  eddy-current  loss  might  remain  the  same? 


GENERATED   VOLTAGES  359 

101.  Total  Core  Loss.  When  the  flux  varies  in  a  mag- 
netic circuit,  there  is  a  loss  due  to  eddy  currents,  and  also, 
as  we  have  seen,  a  loss  due  to  hysteresis.  The  sum  of 
these  two  losses  is  called  the  core  loss  of  the  material. 

We  have  seen  that  the  hysteresis  loss  varies  directly  as 
the  frequency,  but  that  the  eddy-current  loss  varies  as  the 
square  of  the  frequency.  We  have  also  seen  that  the 
hysteresis  loss  varies  as  the  1.6th  power  of  the  maximum  flux 
density,  while  the  eddy-current  loss  varies  as  the  square  of 
the  maximum  flux  density.  These  facts  may  be  summed 
up  in  the  statement 

Ph+e  =  kj  Bnuu1'6  +  fc*f  Bzna*2  Watts,  (28) 

where  k\  and  &2  are  constants. 

These  facts  enable  us  to  predict  how  the  total  core  loss 
in  a  transformer  or  machine  will  vary  with  the  frequency  if 
it  is  changed,  or  with  a  change  of  flux  density.  They 
also  enable  us  to  separate  the  two  losses  and  find  the  value 
of  each  by  methods  which  measure  the  total  loss.  The 
total  core  loss  in  a  machine  can  be  measured  by  finding  the 
power  necessary  to  drive  the  machine  at  no  load  when 
only  these  losses  are  present.  The  core  loss  in  a  transformer 
may  be  measured  by  somewhat  similar  means.  An  ex- 
ample will  show  how  the  losses  can  be  thus  separated. 

Suppose  that  the  total  core  loss  of  a  machine  at  1000 
revolutions  per  minute  is  measured  and  found  to  be  1800 
watts.  Keeping  the  flux  density  constant,  the  machine  is 
then  run  at  1500  revolutions  per  minute,  and  the  total  core 
loss  found  to  be  3000  watts.  It  is  required  to  find  the 
hysteresis  and  eddy-current  losses  separately  at  1000  revo- 
lutions per  minute.  Since  the  flux  density  is  constant 
throughout,  we  may  include  it  with  the  constant  and  write 
our  expression 

P*+e  =  hf  +  A^/2  watts,  (29) 


360      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 
and  inserting  our  values,  we  obtain 

1800  =  /c31000  +  fc410002, 
3000  =  &31500  +  /c415002. 

Since  we  are  dealing  with  a  variation  of  speed  only,  it  makes 
no  difference  whether  our  frequencies  are  expressed  in  cycles 
per  second  or  as  in  this  case  in  revolutions  per  minute. 
Solving  these  two  simultaneous  equations  gives  us 

A*  =  1-4, 

fc4  =  0.0004,  (31) 

and  inserting  these  in  the  expression  for  the  loss  at  1000 
revolutions  per  minute  we  obtain 

1800/,+e  =  1400^  +  400e  watts,  (32) 

showing  that  the  total  loss  of  1800  watts  at  this  speed  is 
divided  into  1400  watts  of  hysteresis  loss  and  400  watts  of 
eddy-current  loss.  If  we  substitute  the  values  in  the 
second  expression,  we  obtain  the  losses  at  1500  revolutions, 
which  gives  us 

3000ft+e  =  2100/,  +  9006  watts,  (33) 

showing  that  at  the  higher  speed,  the  hysteresis  loss  is  2100 
watts  and  the  eddy-current  loss  900  watts. 

Suppose  it  is  required  to  find  the  total  core  loss  in  this 
machine  if  the  flux  density  is  now  increased  by  thirty  per- 
cent. We  will  assume  that  the  speed  is  held  at  1000  revo- 
lutions per  minute. 

The  hysteresis  loss  at  the  new  flux  density  will  be 

Pft  =  1400  X  1.31-6  =  2132  watts,  (34) 

since  the  flux  density  is  1.3  times  as  great  as  before,  and  the 
hysteresis  loss  varies  as  the  1.6th  power  of  the  flux  density. 
The  eddy-current  loss,  similarly,  is 

Pe  =  400  X  1.32  =  676  watts,  (35) 


GENERATED   VOLTAGES  361 

and  hence  we  obtain  for  the  total  core  loss  at  1000  revolu- 
tions per  minute  and  thirty  percent  increase  in  flux  density 

Ph+e  =  2132  +  676  =  2808  watts.  (36) 

Prob.  12-10.  A  transformer  has  a  total  core  loss  of  585 
watts  at  60  cycles  per  second.  The  flux  is  reduced  one-half  and 
the  frequency  is  reduced  to  25  cycles,  and  the  core  loss  becomes 
56.1  watts. 

(a)  What  is  the  eddy-current  loss  at  each  frequency? 

(6)  What  is  the  hysteresis  loss  at  each  frequency? 


SUMMARY    OF   CHAPTER   X 

A  CHANGE  IN  THE  NUMBER  OF   FLUX  LINKAGES 
with  an  electric  circuit  produces  a  voltage  in  the  circuit 


e  =  NIO-8  volts. 
at 

WHEN  MAGNETIC  LINES  OF  FORCE  ARE  CUT  BY  A 
CONDUCTOR  an  electromotive  force  is  set  up  in  the  con- 
ductor. This  is  only  another  way  of  stating  the  above  fact. 
The  same  equation  applies  to  both  ways  of  regarding  the 

d$> 
phenomenon.     In  the  latter  case,  —  means  the  rate  of  cutting 

the  lines  of  force.  IF  THE  THUMB,  FOREFINGER  AND 
MIDDLE  FINGER  of  the  right  hand  are  extended  at  right 
angles  to  one  another,  the  thumb  pointing  in  the  direction  of 
the  motion  of  the  conductor  and  the  forefinger  in  the  direction 
of  the  flux,  the  middle  finger  will  point  in  the  direction  of  the 
generated  voltage.  AN  ALTERNATOR  MAKES  USE  OF  THE 
PRINCIPLE  of  changes  in  flux  linkages  or  the  cutting  of  lines 
of  force  to  set  up  an  alternating  electromotive  force.  BY 
CAUSING  THE  RATE  OF  CUTTING  OF  LINES  OR  CHANG- 
ING IN  FLUX  LINKAGES  TO  VARY  SINUSOID  ALLY,  an 
alternating  electromotive  force  having  a  sine-wave  form  is 
produced. 

e  =  EmaxSin  cot, 
where 

e  =  the  instantaneous  voltage, 

co  =  the  angular  speed  of  the  coil  in  electrical  radians  per 

second, 
t   =  the  time  elapsed  in  seconds. 

BY  ATTACHING  A  SWITCHING  DEVICE  CALLED  A 
COMMUTATOR  to  the  revolving  element  of  an  alternator, 
the  alternating  electromotive  force  produced  in  the  armature 
may  be  delivered  to  the  brushes  as  a  direct  electromotive  force. 
IN  A  HOMOPOLAR  GENERATOR,  the  armature  moves 
through  a  field  of  constant  intensity  and  always  in  the  same 

362 


GENERATED  VOLTAGES  363 

direction,  thereby  inducing  a  direct  electromotive  force.  No 
commutator  is  therefore  needed. 

EDDY  CURRENTS  are  local  electric  currents  which  are 
set  up  in  the  core  and  the  pole  faces  of  machines  or  in  any  solid 
conductor,  when  magnetic  lines  are  cut  by  the  conducting 
materials  composing  these  parts. 

WHEN  THE  CUTTING  OF  FLUX  OR  THE  VARYING  OF 
FLUX  LINKAGES  IS  SINUSOIDAL,  the  eddy-current  loss 
may  be  expressed  by  the  equation 

pav  =  -  h2f2Bmax2  x  iQ-16  watts  per  cubic  centimeter, 
6p 

where 

p  =  the  resistivity  of  the  sheets  in  ohm-centimeters, 
h  =  the  thickness  of  the  sheets  in  centimeters, 
f  =  the  frequency  in  cycles  per  second, 
Bmax  =  the  maximum  flux  density  in  gausses. 

The  energy  of  these  currents  is  dissipated  as  heat  in  the  ma- 
chine. LAMINATING  THE  CORES  decreases  the  eddy 
currents  and  the  accompanying  loss.  THE  TOTAL  CORE 
LOSS  is  the  sum  of  the  hysteresis  and  the  eddy-current  losses. 


PROBLEMS   ON   CHAPTER   X 

Prob.  13-10.  At  what  speed  must  the  rotor  of  Fig.  176 
operate  to  produce  a  60-cycle  frequency? 

Prob.  14-10.  Assume  the  following  data  for  a  48-pole  alter- 
nator with  concentrated  armature  windings. 

Maximum  flux  density  in  conductor  slot,  72,000  lines  per 
square  inch; 

Frequency,  25  cycles; 

Number  of  armature  circuits,  1; 

Number  of  conductors  in  the  armature  circuit,  192; 

Active  length  of  each  conductor,  30  inches; 

Maximum  voltage  desired  per  phase,  3250  volts. 
What  diameter  must  the  rotor  have? 

Prob.  16-10.  What  diameter  would  the  rotor  of  Prob.  14- 
10  have  if  the  frequency  were  60  cycles? 

Prob.  16-10.  The  area  of  a  certain  hysteresis  loop  is  6.21 
square  inches.  The  ordinate  scale  is  5000  gausses  per  inch, 
the  abscissa  scale  is  5  gilberts  per  centimeter  per  inch.  The 
maximum  value  of  the  flux  density  is  16,000  gausses.  What 
is  Steinmetz'  hysteresis  coefficient  for  the  sample?  Frequency, 
60  cycles. 

Prob.  17-10.  The  volume  of  annealed  sheet  steel  in  a  60- 
cycle,  220-2200-volt,  20-kw.  transformer  is  910  cubic  inches 
at  a  maximum  flux  density  of  68,000  lines  per  square  inch. 
The  losses  are  found  to  be  as  follows:  hysteresis  loss  208  watts, 
eddy-current  loss  96  watts.  If  the  frequency  is  changed  to 
25  cycles  and  the  flux  density  to  72,000  lines  per  square  inch, 
what  would  be  the  new  total  loss  and  what  would  be  separately 
the  eddy-current  loss  and  the  hysteresis  loss? 

Prob.  18-10.  An  axle  of  an  ordinary  railroad  coach  has  a 
length  of  8  feet.  The  flux  density  of  the  earth's  field  is  ap- 
proximately 0.6  gauss  and  the  flux  lines  make  an  angle  of  70° 
with  the  horizontal.  The  locomotive  speed  is  70  miles  per 
hour  due  east.  What  voltage  is  generated  in  the  axle?  Does  it 
make  any  difference  whether  the  train  is  running  north  or  east? 

Prob.  19-10.  A  certain  8-pole  generator  has  a  rated  speed 
of  800  revolutions  per  minute.  The  equivalent  maximum 
flux  density  is  65,000  lines  per  square  inch,  The  armature 

364 


GENERATED   VOLTAGES  365 

core  is  made  up  of  "  regular  dynamo  "  steel  sheets  0.014  inch 
thick  and  weighs  1000  pounds.  (Specific  gravity  is  7.79.) 
The  hysteresis  coefficient  is  0.0013  and  the  eddy-current  co- 
efficient is  0.0021.  What  is  the  total  core  loss? 

Prob.  20-10.  To  what  value  would  the  core  loss  of  Prob. 
19-10  be  reduced  if  silicon  steel  sheets  were  substituted  for  the 
"  regular  dynamo  "  sheets?  Specific  gravity,  7.5;  hysteresis 
coefficient,  0.00071;  eddy-current  coefficient,  0.000067. 

Prob.  21-10.  A  solenoid  40  centimeters  long  and  2  centi- 
meters in  diameter  is  wound  with  1200  turns  of  copper  wire. 
The  current  in  the  wire  at  a  certain  instant  is  varying  at  the 
rate  of  400  amperes  per  second.  What  eddy-current  loss  is 
produced  at  this  instant  in  a  round  graphite  rod  of  0.10%  con- 
ductivity, 1  centimeter  in  diameter  and  20  centimeters  long 
placed  centrally  in  the  coil,  with  its  axis  coinciding  with  the 
axis  of  the  coil? 

Prob.  22-10.  What  average  eddy-current  loss  will  take 
place  in  the  rod  of  Prob.  21-10  when  the  coil  carries  an  alter- 
nating current  varying  sinusoidally,  60  cycles  per  second,  with 
a  maximum  value  of  1.24  amperes? 

Prob.  23-10.  A  coil  containing  700  turns  is  wound  on  a 
closed  iron  core  of  5  square  inches  sectional  area  and  15  inches 
effective  length.  Assuming  that  the  flux  in  the  iron  follows 
the  magnetization  curve  for  annealed  sheet  steel:  — 

(a)  Determine  the  constants  of  Froelich's  equation  applied  to 
this  magnetization  curve. 

(6)  Determine  the  equation  showing  the  relation  between 
flux  and  time,  when  the  current  in  the  winding  follows  the  law 

i  =  2  sin  377* ; 

(c)  Bearing  in  mind  that  the  e.m.f.  induced  in  the  coil  is 
given  by  n  ~  »  determine  the  equation  for  the  voltage  induced  in 

the  coil  by  this  flux. 

(d)  Plot  a  half  wave  for  parts  (6)  and  (c)  by  choosing  7  con- 
venient values  of  377  X  t. 

Prob.  24-10.  The  coil  in  Prob.  23-10  has  applied  to  its 
terminals  a  sinusoidal  voltage  of  283  volts  maximum  value. 

(a)  Plot  to  a  suitable  scale  a  half  wave  of  this  voltage. 

(6)  Plot  on  the  same  set  6f  axes  the  flux  through  the  coil 
corresponding  to  this  voltage.  The  flux  equation  may  be  deter- 
mined by  an  integration. 


366      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

(c)  To  each  value  of  flux  there  corresponds  a  definite  mag- 
netizing current.     Plot  the  curve  of  this  current.     This  may 
be  done  by  choosing  values  of  flux  and  obtaining  the  correspond- 
ing values  of  current  from  the  magnetization  curve. 

(If  the  hysteresis  loop  of  the  iron  were  used,  instead  of  the 
magnetization  curves,  the  resulting  current  curve  would  be  dis- 
torted, but  more  nearly  correct.) 

(d)  Compare  the  results  of  Prob.  23-10  and  24-10.     Why 
does  a  sinusoidal  current  correspond  to  a  nonsinusoidal  voltage, 
and  vice  versa? 


CHAPTER  XI 
FORCE  ON  A  CONDUCTOR 

It  is  an  axiom  that  every  electric  generator  is  reversible; 
that  is,  that  if  electric  power  in  the  proper  form  is  supplied 
to  a  machine  used  as  an  electric  generator,  it  will  tend  to  run 
as  an  electric  motor.  In  order  that  such  a  machine  may 
operate  satisfactorily  as  a  motor,  it  is  generally  necessary  to 
make  certain  adjustments  of  brushes,  rheostats  etc.,  and  to 
add  auxiliaries  such  as  starting  motors  and  speed-limiting  de- 
vices, but  the  fact  remains  that  the  tendency  is  present  in 
the  machine  to  operate  as  a  motor.  All  that  is  necessary 
is  the  proper  control  and  regulation  of  this  tendency. 

102.  Force  on  a  Conductor  Carrying  a  Current.  We 
have  seen  that  when  a  conductor  is  moved  sideways  in  a 
magnetic  field,  there  is  a  voltage  generated  in  the  conductor. 
This  principle  gives  rise  to  the  electric  generator.  There  is 
another  principle  which  is  somewhat  similar  and  which 
makes  possible  the  electric  motor.  Whenever  a  conductor 
is  situated  in  a  magnetic  field  and  not  parallel  to  the  flux 
and  is  carrying  an  electric  current,  there  is*  a  force  upon  the 
conductor  which  tends  to  move  it  sideways  through  the 
field. 

Thus  in  Fig.  183,  the  conductor  shown  situated  in  a 
uniform  field  and  carrying  a  current  which  flows  from  back 
to  front  is  acted  upon  by  a  force  which  tends  to  push  it  up. 
In  order  to  find  the  direction  of  a  force  on  a  conductor,  we 
can  use  the  left-hand  rule.  It  will  be  remembered  that  the 
right-hand  rule  was  used  for  a  generated  voltage.  The 
left-hand  rule  is  similar  and  is  used  for  a  motor.  If  the 
thumb  and  first  two  fingers  of  the  left  hand  are  held  per- 
pendicular to  one  another,  the  forefinger  pointing  in  the 

367 


368      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

direction  of  the  flux,  the  middle  finger  pointing  in  the  di- 
rection in  which  the  current  flows  in  the  conductor,  then  the 
thumb  will  point  in  the  direction  in  which  the  force-  tends  to 
move  the  conductor  in  the  field. 


FIG.  183.     A  wire  carrying  a  current  and  situated  in  a  magnetic 
field  is  acted  on  by  a  force. 

The  amount  of  this  force  has  been  found  experimentally 
to  be  proportional  to  the  flux  density,  proportional  to  the 
current  flowing  in  the  conductor  and  proportional  to  the  pro- 
jection of  the  conductor  perpendicular  to  the  field.  If  the 
conductor  is  perpendicular  to  the  field  as  is  shown  in  Fig. 
183,  this  last  factor  becomes  simply  the  length  of  the  con- 
ductor in  the  field.  If  the  conductor  makes  an  angle  0 
with  the  flux  lines,  then  the  length  to  be  taken  is  the  length 
of  the  conductor  multiplied  by  the  sine  of  6.  In  the  fol- 
lowing discussion,  unless  it  is  stated  otherwise,  we  will 
assume  that  the  conductor  is  perpendicular  to  the  field. 

If  we  write  everything  in  c.g.s.  units,  the  force  on  the 
conductor  in  dynes  is  equal  to  the  product  of  the  flux  den- 
sity in  gausses,  the  current  in  abamperes  and  the  length 
of  the  conductor  in  centimeters;  that  is, 

F  =  BlI  dynes.  (1) 

The  reason  that  the  proportionality  factor  is  unity  in  this 
case  may  readily  be  shown. 


FORCE  ON  A   CONDUCTOR  369 

Assume  that  the  above  conductor  is  without  resistance, 
so  that  there  are  no  losses  involved  in  the  flow  of  current 
through  it.  Assume  that  it  is  moving  through  the  field  at  a 
velocity  v.  Then  there  will  be  generated  in  it  a  voltage  of 
amount 

e  =  Blv  abvolts.  (2) 

By  the  right-hand  rule,  referring  to  Fig.  183,  the  direction 
of  this  voltage  will  be  opposite  to  the  direction  in  which 
there  is  flowing  the  current  which  produces  the  force  to 
cause  the  conductor  to  move  with  this  velocity.  That  is, 
the  voltage  is  a  back  voltage.  There  is  hence  an  input  in 
the  conductor  equal  to 

el  =  Blvl  ergs  per  second.  (3) 

This  power  input  is  used  up  in  maintaining  the  conductor 
at  a  velocity  v.  However,  a  force  F  on  a  conductor  which 
is  moving  at  a  velocity  v  in  the  direction  of  the  force  is 
doing  work  at  the  rate 

Fv  ergs  per  second. 

If,  as  we  assumed,  there  is  no  resistance  loss  in  the  conductor, 
and  if  in  addition  we  assume  that  the  velocity  is  constant, 
the  entire  electrical  input  must  appear  as  this  mechanical 
power  output,  in  accordance  with  the  law  of  conservation 
of  energy.  Therefore 

Fv  =  Blvl  ergs  per  second,  (4) 

which  shows  that 

F  =  Ell  dynes.  (1) 

We  thus  see  that  the  force  on  the  conductor  is  equal  to  the 
product  of  B,  I  and  7,  all  in  c.g.s.  units. 
If  /  is  in  amperes,  then 

F  =  £Z      dynes,  (5) 


370      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

If  we  wish  to  convert  this  equation  to  practical  units,  we 
obtain 

,,  _  BlI  1 

=  2.25  X  10-7  BlI  pounds,  (6) 

where 

7  is  in  amperes, 
B  is  in  gausses, 
I  is  in  centimeters. 

If  a  coil  of  a  single  turn,  as  shown  in  section  in  Fig.  184, 
is  situated  in  a  uniform  magnetic  field,  there  will  be  a  torque 

acting  upon  the  coil  which 
will  be  proportional  to  the 
current.     We  will  compute 
the  torque  on  the  assump- 
tion that  the  current  carried 
is    7   abamperes,   that   the 
plane  of  the  coil  makes  an 
angle  0  with  the  plane  of 
the  flux,  that  the  length  of 
FIG.  184.    A  coil  carrying  a  current     the    coil    in    the    direction 
and  situated  in  a  parallel  magnetic     perpendicular  to  the  flux  is  I 
field  is  acted  on  by  a  torque  pro-  j    ,1         ,1          •  ljLl       f  ,-, 

portional   to   the   cosine   of  the     and   that  the  Wldth  °f  the 
angle  6.  CO"  &  2r. 

By  the  left-hand  rule,  the 

upper  conductor  is  forced  upward  and  the  lower  conductor 
is  forced  downward.  These  forces  are  each  of  amount 

F  =  BlI  dynes. 

The  torque  acting  on  the  coil  is  the  couple  produced  by 
these  forces,  or  the  sum  of  their  moments  about  the  axis. 
Each  conductor  is  acted  upon  by  a  force  of  which  the  com- 
ponent perpendicular  to  the  line  joining  the  conductor  to 
the  axis  is 

F  cos  0  =  BlI  cos  0,  (7) 


FORCE  ON  A   CONDUCTOR  371 

The  total  moment,  or  total  turning  torque  acting  upon 
the  coil,  is  therefore 

T  —  2  Bllr  cos  6  dyne-centimeters,  (8) 

which  is  expressed  in  dynes  at  one-centimeter  lever  arm  if 
all  of  the  quantities  in  the  equation  are  in  c.g.s.  units. 
This  may  readily  be  changed  to  be  in  terms  of  pounds  at  a 
lever  arm  of  one  foot. 

Note  that  the  turning  moment  is  at  maximum  when  the 
plane  of  the  coil  is  parallel  to  the  flux  lines;  also  that  when 
the  angle  0  is  ninety  degrees  the  torque  becomes  zero.  If 
6  is  greater  than  ninety  degrees,  the  torque  is  reversed  and 
the  coil  tends  to  turn  backwards.  This  may  be  summed 
up  by  stating  that  the  coil  tends  to  move  into  such  a  position 
that  it  will  link  a  maximum  amount  of  flux  in  a  right-handed- 
screw  direction  with  respect  to  the  current. 

Prob.  1-11.  The  flux  density  in  the  air  gap  of  a  magnet  such 
as  is  shown  in  Fig.  183  is  63,000  lines  to  the  square  inch.  What 
force  acts  on  a  conductor  of  length  4  inches,  carrying  a  current 
of  40  amperes  and  placed  in  this  field  at  an  angle  of  30°  with  the 
flux  lines? 

Prob.  2-11.  What  torque  in  gram-centimeters  acts  on  a  coil 
of  a  single  turn  situated  with  respect  to  a  magnetic  field  as 
shown  in  Fig.  184,  if  the  current  flowing  is  50  amperes,  the 
flux  density  is  7000  gausses  and  r  is  2  centimeters?  Compute 
for  three  values  of  the  angle  6  of  30°,  60°  and  90°.  The  length 
of  each  side  is  12  centimeters. 

Prob.  3-11.  If  the  coil  of  Prob.  2-11  has  4  turns  and  the 
flux  density  is  58,000  lines  to  the  square  inch,  what  will  be  the 
torque  in  pound-inches  when  0=0°  and  the  wire  carries  a 
current  of  30  amperes? 

103.  Meters.  Electric  meters  are  of  many  kinds.  Or- 
dinary direct-current  meters  usually  depend  for  their  action, 
however,  simply  upon  the  force  exerted  upon  a  conductor 
carrying  current  when  situated  in  a  magnetic  field.  This 


372      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

force  is  used  to  deflect  a  pointer  and  thus  gives  a  measure 
of  the  amount  of  current  flowing. 

Portable  instruments  of  this  sort  are  usually  constructed 
as  shown  in  Fig.  185.  The  cross-section  of  the  core  and  poles 
is  shown  in  Fig.  186.  M i  and  M2  are  the  poles  of  a  per- 


FIG.  185.     Cutaway  view  of  a  direct-current  meter. 
Electrical  Instrument  Co. 


Weston 


manent  magnet  which  is  constructed  in  such  a  manner  as 
to  have  a  very  constant  value  of  field  strength.  C  is  a  soft- 
iron  core  cylindrical  in  form  and  situated  between  the  poles. 


FORCE  ON  A  CONDUCTOR  373 

The  core  is  stationary.    A  rectangular  coil  H  as  shown  in  Fig. 
185  is  pivoted  in  bearings  in  such  a  manner  that  its  sides  re- 
volve in  the  space  between  the  magnet  poles  and  the  core. 
The  spindle  on  which  this  coil  is  mounted  carries  a  pointer  P 
which  moves  over  a  scale  and 
gives    the    indication.       The 
pointer  is  returned  to  its  zero 
position  by  means  of  a  hair- 
spring S,  through  which  the 
current   is   also   usually    con- 
ducted to  the  coil.    Sometimes 
two  hairsprings  are  used,  one     FIG.  186.    Cross-section  of  the 
to  conduct  the  current  in  and        core  and  poles  of  the  meter  of 
the  other  to  conduct  the  cur-        Fig.  185.    The  flux  is  radial, 
rent  out  of  the  coil.     One  of 
these  may  be  replaced  by  a  flexible  lead. 

As  shown  in  Fig.  186,  the  influence  of  the  core  C  is  to 
cause  the  flux  to  be  radial  in  the  air  gap  and  of  very  nearly 
constant  flux  density.  The  coil  H  is  therefore  always 
situated  in  a  field  of  the  same  strength.  The  force  acting 
upon  the  sides  of  the  coil  is  simply  proportional  to  the  cur- 
rent flowing  in  the  coil  and  is  independent  of  the  position 
of  the  coil.  The  restoring  force  of  the  hairspring  is  propor- 
tional to  the  deflection.  Thus  the  deflection  of  the  coil, 
and  hence  the  movement  of  the  pointer,  will  be  proportional 
to  the  current  flowing.  This  gives  a  meter  with  evenly 
divided  divisions  on  the  scale. 

It  should  be  noted  that  in  this  construction  we  do  not  have 
a  case  of  a  coil  situated  in  a  parallel  uniform  field.  The  field 
is  everywhere  radial,  so  that  the  force  acting  upon  the  sides 
of  the  coil  is  always  perpendicular  to  the  plane  of  the  coil. 
The  entire  force,  therefore,  and  not  a  component  of  the 
force,  is  effective  in  producing  a  turning  moment. 

As  has  been  noted  before,  such  a  meter  may  be  used  to 
measure  either  voltage  or  current,  depending  upon  the 


374      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

resistance  to  which  the  coil  is  wound.  In  voltmeters,  it  is 
usual  to  supply  an  external  resistance  in  addition  to  the 
resistance  of  the  coil  itself.  Ammeters  are  also  constructed 
by  supplying  an  external  resistance  in  parallel  with  the  coil. 

If  such  a  meter  is  used  in  an  alternating-current  circuit  of 
the  usual  frequencies,  it  will  not  deflect  at  all.  The  current 
in  an  alternating-current  circuit  flows  first  in  one  direction 
and  then  in  the  other,  reversing  its  direction  many  times  a 
second.  When  the  current  is  flowing  in  one  direction,  the 
coil  of  the  meter  tends  to  turn  clockwise,  but  this  is  offset 
by  an  exactly  equal  counter-clockwise  impulse  when  the  cur- 
rent next  reverses.  Such  a  meter,  therefore,  cannot  be 
used  to  measure  alternating  current.  In  order  to  make  an 
alternating-current  meter,  we  replace  the  permanent  mag- 
net by  an  electromagnet,  and  excite  this  electromagnet 
by  a  coil  which  is  connected  in  series  with  the  movable  coil. 
The  direction  of  the  current  in  the  coil,  therefore,  as  well  as 
the  direction  of  the  flux  in  which  it  is  situated,  reverses 
each  half  cycle.  Since  both  the  flux  and  the  current  reverse, 
it  will  easily  be  seen,  by  utilizing  the  left-hand  rule,  that 
the  turning  moment  on  the  coil  is  in  the  same  direction  as 
before.  Such  instruments  will  therefore  deflect  when  an 
alternating  current  flows  through  them  and  may  be  used  to 
measure  alternating  current.  They  are  called  dynamometer 
instruments. 

A  dynamometer  instrument  will  not,  however,  be  an 
instrument  utilizing  an  evenly  divided  scale.  The  flux  in 
which  the  coil  is  situated  is  not  constant  as  in  a  direct- 
current  instrument,  but  depends  upon  the  strength  of  the 
current,  and  in  a  properly  constructed  instrument  is  propor- 
tional to  the  strength  of  the  current.  The  turning  moment 
on  the  coil,  therefore,  which  is  proportional  to  the  product 
of  B  and  7,  is  proportional  to  the  square  of  the  current. 
If  the  current  through  such  an  instrument  is  doubled,  the 
deflection  obtained  will  be  four  times  as  great,  and  so  on. 
The  scale  used  must  be  divided  accordingly.  A  dyna- 


FORCE  ON  A   CONDUCTOR 


375 


Phosphor  Bronze 
Filament 


I  ./Permanent 


Magnet 


mometer  instrument  may  also  be  used,  of  course,  to  measure 
a  direct  current. 

Alternating-current  meters  are  also  made  which  depend 
upon  entirely  different  principles,  such  as  the  heating  pro- 
duced in  a  wire,  the  tendency  of  a  piece  of  soft  iron  to  line 
up  with  a  field,  and  so  on. 

The  same  construction  as  is  used  in  the  direct-current 
meter  may  be  used  also  in 
constructing  a  sensitive  gal- 
vanometer. In  this  case  the 
coil,  instead  of  being  pivoted, 
is  usually  suspended  on  a  fine 
suspension  thread  as  shown 
in  Fig.  187.  This  suspended 
coil  often  carries  a  small  mirror 
by  which  its  deflection  may 
be  noted  by  reflecting  a  spot 
of  light  from  the  mirror  onto 
a  scale,  or  by  looking  at  the 
reflection  of  a  scale  in  the 
mirror  as  seen  through  a  tele- 
scope. The  principle  of  opera- 
tion is  exactly  the  same  as 
that  of  a  direct-current  meter. 
By  making  a  coil  of  a  large 

number  of  turns  of  fine  wire,  by  using  a  strong  mag- 
netic field  and  by  using  a  very  delicate  suspension,  the 
galvanometer  may  be  made  to  deflect  when  extremely  small 
values  of  current  are  used. 

A  galvanometer  of  long  period  is  called  a  ballistic  gal- 
vanometer, and  is  used,  as  we  have  seen,  for  the  measure- 
ment of  flux.  The  deflection  of  a  ballistic  galvanometer  is 
proportional  to  the  total  quantity  which  flows  through  it; 
that  is,  it  is  proportional  to 


FIG.  187.     Diagram  of  a  sensi- 
tive galvanometer. 


q  =  /  idt  abcoulombs, 


(9) 


376      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

which,  if  we  assume  for  the  moment  that  we  can  neglect  the 
inductance  of  the  galvanometer,  is  equal  to 


-  I  edt.  (10) 

We  have  seen  in  a  previous  chapter,  however,  that  the 
total  amount  of  flux  linking  a  measuring  coil  can  be  found 
from  the  expression 

(f>  =  I  edt  maxwells,  (11) 

and  so  the  total  amount  of  flux  can  be  found  from  the  de- 
flection of  a  ballistic  galvanometer  connected  to  the  meas- 
uring coil. 

The  reason  why  the  deflection  of  a  ballistic  galvanometer 
is  proportional  to  q,  that  is,  to  the  quantity  passing  through 
it,  is  as  follows.  The  force  acting  on  the  coil  is  proportional 
to  the  current,  since  the  coil  is  situated  in  a  field  of  constant 
flux  density.  The  acceleration  of  the  coil  is  proportional  to 
this  force;  that  is, 

*jL=KiF=  K2i,  (12) 

where  K\  and  KI  are  proportionality  factors.  The  initial 
velocity  acquired  by  the  coil  is  therefore 

tt  (13) 

which  we  have  seen  is  proportional  to  the  flux.  The  de- 
flection of  the  coil  depends  simply  upon  the  initial  velocity 
given  to  it,  just  as  the  deflection  of  a  ballistic  pendulum  used 
to  measure  the  velocity  of  a  bullet  depends  simply  upon  the 
initial  velocity  imparted  to  it.  The  deflection  of  the  bal- 
listic galvanometer  which  is  proportional  to  its  initial 
velocity  will  thus  be  proportional  also  to  the  total  number 
of  flux  linkages  changed  in  the  measuring  coil. 
Ill  this  expression  we  have  neglected  the  inductance  of 


FORCE  ON  A   CONDUCTOR  377 

the  coil  itself.  It  will  be  seen,  however,  in  the  solution  of 
Prob.  4-11,  that  the  quantity  of  electricity  passed  through 
an  inductive  circuit,  when  a  constant  voltage  is  applied  to 
it  for  a  given  interval  of  time  and  then  removed,  is  the  same 
as  the  quantity  which  would  pass  if  the  inductance  were  not 
present.  To  be  sure,  the  current  would  take  a  longer  time 
in  reaching  its  maximum  value,  but  it  would  also  persist 
enough  longer  to  make  the  total  quantity  of  electricity 
passed  through  the  circuit  exactly  the  same.  The  induc- 
tance of  the  coil  of  the  ballistic  galvanometer,  therefore,  does 
not  affect  the  proportionality  of  its  readings  to  the  flux  to 
be  measured.  We  also  note  that  it  makes  no  difference 
whether  the  measuring  coil  has  its  voltage  induced  in  it 
for  a  short  or  a  long  period,  provided  the  total  change  of 
flux  linkages,  that  is, 

Cedt, 

remains  unchanged.  This  is  strictly  true  only  if  the  meas- 
uring coil  is  removed  from  the  influence  of  the  flux  in  a  time 
sufficiently  short  so  that  the  coil  of  the  ballistic  galvanom- 
eter has  not  appreciably  moved  from  its  position  in  the  mean- 
while. In  order  to  make  sure  that  this  condition  is  fulfilled, 
the  coil  is  made  with  a  large  moment  of  inertia,  resulting 
in  a  long  period. 

Prob.  4-11.  Prove  the  statement,  "  When  a  constant  volt- 
age is  applied  to  an  inductive  circuit  for  a  given  length  of  time 
and  then  removed  without  opening  the  circuit,  the  same  quan- 
tity of  electricity  passes  as  would  pass  in  a  non-inductive  cir- 
cuit if  the  same  voltage  were  applied  for  the  same  length  of 
time,  the  resistances  of  the  two  circuits  being  equal." 

104.  Motors.  A  coil  of  wire  carrying  current  and  situated 
in  a  magnetic  field  tends  to  turn  about  an  axis  perpendicular 
to  the  field,  and  so  may  be  used  as  a  motor  and  caused  to 
perform  useful  work.  Referring  to  Fig.  188,  in  which  a 
single-turn  coil  is  shown  for  convenience,  we  see  by  the  left- 
hand  rule  that  the  coil  tends  to  turn  in  a  clockwise  direction 


378      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

when  a  current  flows  in  the  coil  in  the  direction  indicated  by 
the  dot  and  cross. 

When  this  coil  has  arrived  at  a  position  such  that  its  plane 
is  perpendicular  to  the  flux  lines,  however,  the  force  acting 


FIG.  188.     A  current  in  the  direction  indicated  tends  to  rotate  the 
coil  in  the  direction  of  the  arrow. 

on  the  wire  will  be  in  the  plane  of  the  coil  and  the  turning 
moment  about  the  axis  will  be  zero.  In  fact,  if  the  coil  were 
turned  beyond  this  position  to  a  position  as  shown  by  the 
dotted  lines,  its  turning  moment  would  be  reversed  and  it 
would  tend  to  turn  backward. 

Suppose,  however,  that  just  as  the  coil  of  Fig.  188  passes 
through  the  vertical  position,  we  reverse  the  direction  of 
the  current  in  the  coil.  After  passing  this  dead  center,  it 
will  take  the  position  of  Fig.  189,  and  since  the  current  is 


FIG.  189.  When  the  coil  of  Fig.  188  has  reached  this  position,  the  cur- 
rent is  reversed  and  it  tends  to  continue  rotating  in  the  same 
direction. 

reversed,  the  turning  moment  on  the  coil  will  be  in  the  same 
direction  as  before,  so  that  it  will  tend  to  keep  on  turning. 
By  reversing  the  current  in  the  coil  at  the  correct  intervals, 
we  are  thus  able  to  render  the  turning  moment  always  in 
the  same  direction  and  so  construct  a  motor  which  will 
keep  turning  continuously. 


FORCE  ON  A   CONDUCTOR  379 

One  way  in  which  this  can  be  accomplished  is  by  sup- 
plying the  coil  with  an  alternating  current.  If  the  rotation 
of  the  coil  then  simply  keeps  step  with  the  alternation  of 
the  current  supplied,  that  is,  if  the  armature  of  the  motor 
runs  hi  synchronism  with  the  supply,  the  torque  will  always 
be  in  the  correct  direc- 
tion and  the  motor  fl4 
will  continue  to  re- 
volve. Such  an  al- 
ternating-current mo- 
tor is  called  a  syn-  3  \  /  Time 
chronous  motor.  In 
Fig.  190  is  shown  the  FlG-  190-  Showing  the  relation  between 
manner  in  which  the  the  current  in  the  coil  and  the  position  of 
,  •  ,  r  the  coil.  Note  that  the  current  is  zero 
at  the  instant  when  a  current  would 
mature,  that  is,  in  the  produce  no  torque, 
revolving  coil,  reverses 

with  the  time,  and  over  various  points  in  the  wave  are 
shown  for  convenience  the  succeeding  positions  of  the  revolv- 
ing coil. 

In  such  a  motor,  the  coil  is,  of  course,  actually  imbedded 
in  iron,  just  as  in  a  generator,  in  order  to  reduce  the  reluc- 
tance of  the  flux  path.  Synchronous  motors  are  usually  made 
multipolar  and  generally  polyphase;  this  means  that  there  are 
several  coils  with  different  time  relations  operating  simul- 
taneously on  one  armature.  Synchronous  motors  as  well  as 
alternators  are  also  usually  built  with  revolving  fields  instead 
of  revolving  armatures.  Since  action  and  reaction  are  always 
equal  and  opposite,  it  makes  no  difference  in  the  amount  of 
torque  produced  whether  the  armature  or  the  field  revolves. 

Alternating-current  motors  are  also  made  in  various 
other  forms  such  as  the  induction  motor  or  the  repulsion 
motor.  The  principles  of  these  are  too  complicated  to  be 
discussed  at  this  time. 

Instead  of  reversing  the  direction  of  the  current  supply  to 
a  motor  as  the  armature  revolves,  we  may  use  a  commutator 


380      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

to  reverse  the  connections  of  the  armature  conductors  to 
the  external  circuit  and  thus  construct  a  direct-current  mo- 
tor. In  fact,  any  direct-current  generator  will,  if  supplied 
with  current,  run  as  a  direct-current  motor. 

Examine  again  the  arrangement  shown  in  Fig.  168, 
which  was  drawn  for  a  direct-current  generator.  Suppose 
that  the  current  enters  by  brush  A  and  leaves  by  brush  B. 
By  tracing  the  connections  through  the  winding,  we  find  that 
the  current  goes  down  through  each  of  the  conductors  under 
the  south  pole  and  comes  up  through  each  of  the  conductors 
under  the  north  pole.  The  machine  will  thus  tend  to  turn 
in  a  clockwise  direction,  that  is,  in  a  direction  exactly  reverse 
to  the  direction  in  which  it  ran  as  a  generator.  Suppose 
that  the  armature  turns  through  one-sixth  of  a  revolution. 
Again  examine  the  path  of  the  current.  It  is  found  as 
before  that  the  current  goes  down  through  each  of  the  con- 
ductors situated  under  the  south  pole  and  up  through  each 
of  the  conductors  on  the  other  side  of  the  armature.  What- 
ever the  position  of  the  armature,  therefore,  the  torque 
remains  in  the  same  direction  and  of  approximately  the 
same  amount,  so  long  as  the  current  entering  through  the 
brush  is  constant.  This  machine  will  hence  run  as  a  motor. 

The  homopolar  machine  shown  in  Fig.  178  will  also  run 
as  a  motor  if  supplied  with  current.  In  this  case  the  con- 
ductor is  always  situated  in  a  field  of  the  same  strength  and 
direction,  no  matter  what  the  position  of  the  armature.  The 
torque  is  consequently  constant  in  magnitude  and  direction, 
so  long  as  the  current  through  the  conductor  remains  un- 
changed. Machines  of  this  sort  are  not,  however,  used  as 
motors,  for  they  are  subject  to  strict  practical  limitations 
and  for  most  purposes  would  be  far  too  expensive  to  be 
considered. 

105.  The  Back  Electromotive  Force,  The  force  acting 
on  a  conductor  in  a  magnetic  field  is  proportional  to  the 
strength  of  the  field  and  to  the  current  in  the  conductor. 
The  force  remains  unchanged  when  the  conductor  moves. 


FORCE  ON  A  CONDUCTOR  381 

It  makes  no  difference  whether  the  conductor  is  at  rest  or  is 
traveling  at  a  high  velocity,  the  force  is  independent  of 
the  speed  of  the  conductor  and  depends  simply  upon  the 
current  and  the  field  strength.  In  other  words,  in  a  motor 
the  torque  is  proportional  to  the  total  flux  and  to  the  motor 
current,  and  is  independent  of  the  speed  of  the  motor. 

In  the  same  way,  a  conductor  which  is  moving  through  a 
magnetic  field  has  a  voltage  generated  in  it  which  is  pro- 
portional to  the  strength  of  the  field  and  to  the  velocity  of 
the  conductor.  It  has  the  same  voltage  generated  in  it 
whether  a  current  flows  in  the  conductor  or  not.  The 
terminal  voltage  across  the  conductor  may  depend  upon  the 
current  flowing,  for  the  terminal  voltage  is  the  combination 
of  the  generated  voltage  and  the  resistance  drop.  The 
voltage  generated,  however,  depends  simply  upon  the  field 
strength  and  the  velocity  of  the  conductor.  That  is,  in  a 
dynamo  there  is  a  certain  generated  voltage  which  is  pro- 
portional to  the  speed  of  the  machine  and  to  the  total  flux, 
and  which  in  no  way  depends  upon  the  current  flowing. 

Suppose  that  we  have  a  motor  which  is  carrying  a  current 
/,  and  that  the  terminal  voltage  of  the  motor  is  E.  The 
total  power  input  into  the  motor  is  IE.  A  portion  of  this 
input,  PR,  is  converted  into  heat  in  the  motor.  In  this 
case  R  is  the  resistance  of  the  windings,  on  the  armature. 

The  remainder  of  the  input, 

P  =  El  -  PR  watts,  (14) 

is  converted  into  mechanical  work  in  the  motor.  The  motor 
is  a  machine  for  transforming  electrical  power  into  me- 
chanical power.  The  amount  of  power  thus  transformed  is, 
of  course,  equal  to  the  power  input  to  the  machine  minus  the 
power  losses  in  the  machine  itself.  Of  the  amount  of  power 
transformed  into  mechanical  form,  a  portion  will  naturally 
be  lost  in  the  friction  of  the  bearings  and  so  on,  and  hence 
will  be  unavailable  for  useful  purposes. 
This  motor,  however,  must  also  act  as  a  generator,  when 


382      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

it  is  running.  It  has  generated  in  it  a  voltage  which  is 
proportional  to  the  speed  at  which  it  runs.  This  voltage, 
we  have  seen,  is  in  a  direction  opposite  to  the  direction 
of  the  applied  voltage,  that  is,  it  is  a  back  electromotive 
force.  We  will  denote  it  by  EB.  The  current  flowing 
through  the  motor  is  determined  by  Ohm's  law;  that  is,  the 
current  is  equal  to  the  electromotive  force  acting  divided 
by  the  resistance  of  the  motor  winding.  In  applying  this 
law,  however,  we  must  take  into  account  all  of  the  voltages 
acting.  There  is  in  addition  to  the  applied  voltage  a  back 
electromotive  force  which  opposes  this  voltage.  The  net 
voltage  acting  is  therefore 

E  —  EB, 

and  hence  the  current  is 

/  =  E  ~EB  amperes.  (15) 

K 

We  may  write  this  last  expression  as 

E  =  EB  +  IR  volts.  (16) 

If  we  insert  it  into  the  expression  above  for  the  amount  of 
power  which  is  transformed  into  mechanical  form,  we  shall 
obtain 

P  =  (EB  +  IR)  I  -  PR  =  EBI  watts.  (17) 

The  amount  of  power  transformed  by  a  motor  from  elec- 
trical to  mechanical  form  is  thus  equal  to  the  current  times 
the  back  electromotive  force.  The  output  of  the  motor  will 
be  equal  to  the  amount  of  power  transformed  minus  the 
frictional  losses. 

We  may  define  the  efficiency  of  a  machine  as  the  output 
divided  by  the  input;  that  is, 

~  .  output  /10X 

efficiency  =  ~ — *-—  >  (18) 

input 

or 

—  .  input  —  losses 

efficiency  =          . • 

input 


FORCE  ON  A   CONDUCTOR  383 

In  the  losses  must  be  included  the  frictional  losses  in  the 
machine  and  the  electrical  losses  in  the  form  of  PR  in  the 
winding. 

Prob.  5-11.  A  motor  takes  from  the  line  48  amperes  at 
110  volts.  The  PR  losses  are  6%  of  the  input.  What  is  the 
power  transformed  into  mechanical  form?  The  frictional  losses 
are  8%  of  the  input.  What  is  the  output  of  the  motor  in 
horse  power? 

Prob.  6-11.  A  motor  running  at  550  volts  takes  an  arma- 
ture current  of  26  amperes  at  a  certain  load.  The  armature 
resistance  is  1.6  ohms.  What  is  the  resistance  loss  and  what 
is  the  back  electromotive  force?  How  much  power  is  trans- 
formed into  mechanical  form? 

Prob.  7-11.  What  is  the  efficiency  of  the  motor  of  Prob. 
6-11  if  in  addition  to  the  I2R  loss  in  the  armature  there  is 
allowed  630  watts  for  friction,  windage  and  all  other  losses? 

106.  Speed.  The  back  electromotive  force  of  a  motor  is 
proportional  to  its  speed  and  to  the  total  flux;  that  is, 

EB  =  KS<f>  volts,  (19) 

where  K  is  simply  a  proportionality  factor.  We  have  seen, 
however,  that 

E  =  EB  +  IR  volts,  (20) 

and  if  we  substitute  in  this  equation  for  EB  its  value  above, 
we  obtain 

E  =  KS<t>  +  IR  volts,"  (21) 

which,  solved  for  S,  gives 

1       ET   rr> 

S  =  -~  -          -  revolutions  per  minute.  (22) 

A  <p 

This  is  a  very  important  equation  indeed,  since  it  shows  the 
speed  at  which  a  motor  will  run. 

Let  us  examine  first  the  speed  at  which  a  motor  will  run 
under  very  light  load.  In  this  case,  the  output  is  small, 
and  accordingly  the  term  IR  must  be  small,  for  the  current 
is  small.  Neglecting  this  term,  we  obtain 

E  =  KS<j>  volts. 


384      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

This,  however,  is  exactly  the  expression  which  we  obtained 
for  the  back  electromotive  force.  In  other  words,  under 
very  light  loads  a  motor  will  run  at  such  a  speed  that  the 
back  electromotive  force  approximately  equals  the  applied 
electromotive  force.  This  must,  of  course,  be  true  only 
when  a  small  current,  that  is,  a  small  input,  is  to  flow  into  the 
machine.  Suppose  that  we  have  a  machine  which,  when 
run  as  a  generator  at  1000  revolutions  a  minute,  delivers 
a  voltage  of  220  volts.  Let  us  maintain  the  flux  of  this  ma- 
chine strictly  constant,  and  connect  it  to  a  supply  giving 
220  volts,  in  order  to  run  it  as  a  motor.  If  it  is  lightly  loaded, 
it  will  run  at  approximately  the  same  speed  as  before,  that  is, 
at  a  speed  slightly  less  than  1000  revolutions  per  minute. 

When  a  motor  is  loaded,  the  current  will  increase,  for  of 
course  its  input  must  increase.  In  the  speed  equation, 
therefore,  the  term  IR  increases  and  the  speed  accordingly 
drops.  A  constant-flux  motor  will  thus  slow  down  slightly 
as  the  load  is  applied.  It  will  not  slow  down  much,  for  it 
is  necessary  to  decrease  the  speed  only  enough  so  that  the 
difference  between  E  and  EB  is  sufficient  to  force  the  load 
current  through  the  small  internal  resistance  of  the  motor. 

Suppose,  for  instance,  that  the  above  motor  is  loaded  until 
its  input  is  440  watts,  that  is,  until  the  current  is  2  amperes. 
If  the  internal  resistance  of  the  machine  is  4  ohms,  the 
term  IR  in  the  speed  equation  will  be  8  volts.  The  speed 
under  this  load  will  therefore  be 

1   220-8 


and  comparing  this  with  the  speed  at  no  load 

1   220 

^    ^"T1 

we  see  that  the  speed  is  decreased  in  the  ratio 

&     '212. 
So      220' 


FORCE  ON  A   CONDUCTOR  385 

this  means  that  the  speed  decreases  about  four  percent  when 
this  load  is  applied.  This  assumes  that  the  value  of  <f>  will 
remain  unchanged  when  the  armature  current  is  increased. 
In  most  machines  there  is  a  slight  change  in  the  value  of 
<£  from  no-load  to  full-load  conditions  which  introduces  a 
small  error  in  equation  (26). 

Referring  again  to  the  speed  equation,  we  see  that  the 
speed  at  which  the  motor  will  run  is  approximately  pro- 
portional to  the  applied  voltage  so  long  as  the  flux  remains 
constant. 

Moreover,  with  a  constant  applied  voltage,  the  speed  is 
inversely  proportional  to  the  flux.  This  is  again  an  ap- 
proximate relation.  If  the  flux  is  doubled,  the  speed  will 
be  halved.  In  other  words,  with  the  increased  flux,  the 
motor  will  need  to  run  only  half  as  fast  in  order  to  generate 
a  back  electromotive  force  approximately  equal  to  the 
applied  voltage.  Conversely,  if  the  flux  is  halved  in  value, 
the  speed  of  the  motor  will  be  approximately  doubled.  It 
will  be  necessary  for  the  motor  to  run  twice  as  fast  as  before 
in  order  that  its  back  electromotive  force  may  approxi- 
mately balance  the  voltage  of  the  line  and  prevent  a  very 
heavy  current  from  flowing. 

Suppose  that  we  have  a  motor  in  which  the  flux  is  ob- 
tained by  a  magnetizing  coil  connected  directly  across  the 
line.  This  form  of  motor  is  known  as  a  shunt  motor,  and  the 
magnetizing  coil  is  known  as  the  field  winding.  In  order 
to  change  the  speed  of  the  motor,  we  need  simply  to  adjust 
the  current  in  the  field  winding.  If  we  decrease  this  cur- 
rent, the  speed  will  increase;  if  we  increase  the  field  current, 
the  speed  will  be  lowered. 

Prob.  8-11.  A  220- volt  motor  has  a  no-load  speed  of  1200 
revolutions  per  minute.  When  fully  loaded,  the  armature 
carries  92  amperes  and  the  field  flux  decreases  6%.  The  arma- 
ture resistance  is  0.057  ohm.  What  is  the  full-load  speed? 

Prob.  9-11.  If  an  extra  winding  were  put  on  the  fields  of 
the  motor  of  Prob.  8-11  so  that  the  field  flux  increased  6% 
from  no  load  to  full  load,  what  would  be  the  full-load  speed? 


SUMMARY    OF   CHAPTER   XI 

A  CONDUCTOR  CARRYING  AN  ELECTRIC  CURRENT 
and  situated  in  a  magnetic  field  (but  not  parallel  to  it)  is  acted 
upon  by  a  force  according  to  the  equation 

F  =  Ell  dynes, 

where  I  =  length  perpendicular  to  flux. 

EXTEND  THE  THUMB  AND  FIRST  TWO  FINGERS  OF 
THE  LEFT  HAND  perpendicular  to  one  another,  the  middle 
finger  pointing  in  the  direction  of  the  electric  current,  the  fore- 
finger in  the  direction  of  the  flux,  and  the  thumb  will  indicate 
the  direction  in  which  the  conductor  tends  to  move. 

In  a  DIRECT-CURRENT  METER,  the  moving  coil  is  usually 
situated  in  a  constant,  uniform,  radial  field.  The  moving  force 
is  therefore  proportional  to  the  current  in  the  coil.  This 
current  can  be  made  proportional  either  to  the  current  or  to 
the  voltage  it  is  desired  to  measure. 

In  an  ALTERNATING-CURRENT  METER,  the  field  as  well 
as  the  current  in  the  coil  is  proportional  to  the  voltage  or  to 
the  current  to  be  measured.  The  force  on  the  coil  is  therefore 
proportional  to  the  square  of  these  quantities. 

A  BALLISTIC  GALVANOMETER  can  be  made  to  deflect 
proportionally  to  the  quantity  of  electricity  passing  through  it. 

A  MOTOR  OWES  ITS  TORQUE  to  the  force  on  a  con- 
ductor carrying  a  current  in  a  magnetic  field. 

WHEN  THE  ARMATURE  OF  A  MOTOR  ROTATES,  an 
electromotive  force  is  induced  in  the  winding  which  opposes 
the  flow  of  the  current  driving  the  motor.  This  electromotive 
force  is  therefore  called  a  BACK  electromotive  force. 

THE  CURRENT  TAKEN  BY  THE  ARMATURE  OF  A 
MOTOR  can  be  found  from 

T      E  ~EB 

TET 

THE  SPEED  OF  A  MOTOR  IS  DIRECTLY  PROPOR- 
TIONAL to  the  back  electromotive  force  and  inversely  pro- 
portional to  the  field  flux. 


S  = 


1  E  -  IRa 

K  "     0 
386 


PROBLEMS   ON   CHAPTER   XI 


Prob.  10-11.  A  Weston  5-15-150-volt  voltmeter  has  a 
movable  coil  of  the  dimensions  shown  in  Fig.  191.  The  wind- 
ing consists  of  a  single  layer  of  75  turns  of  wire.  For  a  half- 
scale  deflection,  the  current  in  the  coil  is  0.009  ampere.  What 
torque  is  exerted  in  this  coil  if  the  flux  density  in  the  air  gap 
at  all  points  is  1000  gausses?  (The  flux  lines  are  radial:  note 
figure.) 

Prob.  11-11.  Another  type 
of  Weston  meter  has  the  same 
coil  dimensions  as  in  Fig.  191. 
The  gap  flux  density  is  1000 
gausses  and  the  coil  has  a  single 
layer  of  36  turns.  A  current 
of  0.010  ampere  gives  a  move- 
ment of  3  inches  on  a  scale  3.37 
inches  from  the  axis  of  the  mov- 
ing coil.  What  is  the  torque 
and  what  work  is  done  in  mov- 
ing the  coil  this  distance? 

Prob.  12-11.  A  single  turn 
of  wire  as  in  Fig.  184  carries  a 

current  of  75  amperes.  The  radius  of  the  coil  is  10  centi- 
meters, the  length  is  8  centimeters  and  the  flux  density  in  the  gap 
50  kilolines  to  the  square  inch.  What  work  in  joules  is  done 
in  moving  the  coil  from  6  =  0°  to  0  =  90°?  If  the  coil  moves 
through  this  distance  in  0.005  second,  what  is  the  average  power? 

Prob.  13-11.  What  work  will  be  done  if  the  coil  of  Prob. 
12-11  is  rotated  by  hand  through  360°? 

Prob.  14-11.  In  Prob.  12-11,  what  force  will  be  exerted 
tangent  to  the  surface  of  an  8-inch  pulley  secured  to  the  axis 
of  the  coil  if  6  =  0°? 

Prob.  16-11.  Develop  a  formula  for  the  magnetic  force  in 
pounds  between  the  two  wires  (line  and  return)  of  a  transmission 
line.  Let  the  current  be  in  amperes,  the  length  of  the  line  in 
feet  and  the  distance  between  wires  in  inches. 

387 


FIG.  191.     The  movable  coil  of 
a  Weston  voltmeter. 


388      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Prob.  16-11.  A  two-wire  transmission  line  (line  and  return) 
spaced  18  inches  is  strung  on  poles  50  feet  apart,  (a)  In 
what  direction  is  the  magnetic  force  exerted  by  one  wire  on  the 
other?  (6)  How  great  is  the  force  on  a  50-foot  span  if  each 
conductor  carries  400  amperes? 

Prob.  17-11.  In  one  place,  the  transmission  line  of  Prob. 
16-11  goes  under  a  bridge.  Here  it  is  cleated  to  the  wall  every 

5  feet.  The  conductors  are  spaced  6 
inches  apart,  (a)  What  is  the  mag- 
netic force  on  each  cleat  when  the 
line  carries  the  normal  load  current 
of  400  amperes?  (6)  What  would 
be  the  momentary  force  on  each 
cleat  if  the  line  were  accidentally 
short-circuited  and  there  were  a 
momentary  rush  of  current  of  15,000 
amperes? 

Prob.  18-11.  Fig.  192  represents  a 
disconnecting  switch  on  a  high-tension 
line.  It  requires  a  pull  of  5  pounds 
at  R  to  open  the  switch.  A  and  B 
may  be  considered  to  be  of  indefinite 
length.  The  other  conductors  are 
too  remote  to  affect  the  switch. 
How  many  amperes  must  flow  in  a 
short  circuit  which  forces  the  switch 
open? 

Prob.  19-11.  The  current  in  the  armature  of  a  220-volt 
motor  at  full  load  is  90  amperes.  The  resistance  of  the  arma- 
ture is  0.140  ohm.  The  speed  at  full  load  is  1100  revolutions 
per  minute.  Assume  that  the  flux  decreases  4%  as  the  load 
increases  from  no  load  to  full  load  and  determine  the  no-load 
speed.  Armature  current  at  no  load,  5.0  amperes. 

Prob.  20-11.  What  will  be  the  speed  of  the  motor  in  Prob. 
19-11, 

(a)  At  ^  full-load  current? 
(6)  At  1J  full-load  current? 

Prob.  21-11.  The  value  of  <f>  at  no  load  in  the  motor  of 
Prob.  19-11  is  1,500,000  maxwells.  How  many  active  con- 
ductors are  there  in  each  path  between  the  positive  and  the 
negative  brushes? 


FIG.  192.  A  disconnect- 
ing switch  on  a  high- 
tension  line. 


FORCE  ON  A   CONDUCTOR 


389 


Prob.  22-11.  At  what  speed  must  the  motor  in  Prob.  19-11 
be  run  to  operate  as  a  generator  and  deliver  90  amperes  at 
230  volts?  Assume  that  the  flux  at  no  load  will  be  103%  of  the 
no-load  flux  of  Prob.  19-11,  and  will  decrease  4%  as  the  load  rises 
from  zero  to  90  amperes. 

Prob.  23-11.  In  Fig.  193  are  shown  two  conducting  "  rails  " 
and  two  conducting  "  sliders,"  A  and  B. 


FIG.    193.      Sliders  A  and  B  cut  lines  of  magnetic  flux  as  they  move  in 
the  direction  of  F. 

The  sliders  move  in  frictionless  guides  always  perpendicular 
to  the  center  line  between  the  rails.  They  are  connected  by 
an  insulated  link  as  shown.  The  density  of  the  magnetic  field 
perpendicular  to  the  plane  formed  by  the  rails  and  sliders  is  every- 
where 1200  maxwells  per  square  centimeter. 

Each  rail  has  a  resistance  of  20  microhms  per  centimeter. 
Each  slider  has  a  resistance  of  50  microhms  per  centimeter. 

What  force  in  dynes  will  be  required  at  F  in  the  direction 
shown  to  move  the  sliders  in  that  direction  with  a  velocity  of 
80  centimeters  per  second,  when  slider  A  is  10  centimeters 
from  point  C?  What  will  be  the  tension  or  compression  in  the 
connecting  link?  If  this  force  at  F  is  held  constant,  with 
what  velocity  will  the  sliders  be  moving  when  slider  B  is  at 
point  D?  Neglect  mass  of  sliders  and  self-inductance.  (This 
strictly  hypothetical  problem  is  presented  because  practical 
problems  involving  the  same  principles  contain  so  many  other 
factors  that  they  become  too  complicated  for  our  use.) 

Prob.  24-11.  A  pair  of  busbars,  line  and  return,  spaced  12 
inches  on  centers,  extend  15  feet  vertically  down  a  wall.  At 


390      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

the  foot  of  the  wall  the  busses  extend  out  15  feet  perpendicularly 
to  the  wall.  These  horizontal  sections  are  supported  by  in- 
sulators spaced  5  feet  apart,  the  first  pair  being  directly  under 
the  vertical  parts  of  the  busses.  At  both  ends  the  busbars  are 
attached  to  twin  cables.  Consider  the  busbars  hinged  at  the 
bends  and  treat  the  horizontal  parts  as  "  simple  beams."  If 
the  busbars  carry  6000  amperes, 

(a)  What  vertical  stresses  due  to  this  current  will  be  put 
upon  the  pair  of  insulators  which  are  5  feet  from  the  wall? 

(6)  What  horizontal  stresses  will  be  set  up  in  these  same 
insulators? 

(c)  What  will  be  the  total  added  stress  on  these  insulators? 

Prob.  25-11.  If  the  vertical  and  the  horizontal  sections  of 
the  busses  in  Prob.  24-11  instead  of  being  15  feet  long  were 
so  long  that  they  could  be  considered  infinite  in  length,  what 
would  be  the  answers  to  parts  (a),  (b)  and  (c)  of  that  problem? 

Prob.  26-11.  If  the  busses  of  Prob.  24-11  were  spaced  12 
feet  instead  of  12  inches  on  centers,  what  would  be  the  an- 
swers to  parts  (a),  (b)  and  (c)  of  that  problem? 

Prob.  27-11.  Three  parallel  conductors  are  strung  at  the 
corners  of  an  18-inch  equilateral  triangle.  At  a  given  instant 
conductor  A  is  carrying  100  amperes  in  a  given  direction,  and 
B  and  C  are  carrying  50  amperes  each  in  the  opposite  direction. 
What  is  the  amount  and  direction  of  the  magnetic  force  ex- 
erted at  this  instant  on  100  feet  of  each  conductor? 

Prob.  28-11.  If  the  three  conductors  of  Prob.  27-11  had 
been  strung  in  one  plane  and  the  outside  wires  spaced  18  inches 
from  the  center  wire,  what  would  be  the  answers  to  that  prob- 
lem, if 

(a)  Conductor  A  were  an  outside  wire? 
(6)  Conductor  A  were  the  middle  wire  ? 


CHAPTER  XII 
CONDUCTION  THROUGH  GASES 

In  the  preceding  chapters  we  have  studied  metallic  con- 
duction. This  includes  nearly  all  cases  of  the  conduction 
of  electricity  where  Ohm's  law  is  obeyed.  We  have  seen 
that  in  this  case  every  element  of  a  circuit  has  a  certain 
definite  resistance  which  can  be  expressed  numerically. 
These  resistances  may  be  combined  in  series  and  in  parallel 
in  order  to  solve  networks  of  circuits.  Where  several  sources 
of  electromotive  force  are  involved,  Kirchhoff's  laws  may 
be  applied  in  order  to  arrive  at  the  solution. 

Metallic  conduction,  we  saw,  takes  place  by  reason  of 
the  movement  of  free  electrons  through  the  body  of  a 
metal.  These  electrons  in  large  numbers  are  present  in 
the  material  and  free  to  move  through  it,  although  the 
molecules  of  the  material  are  fixed  in  position.  Upon  the 
application  of  an  electromotive  force,  however  small,  the 
electrons  wander  towards  the  region  of  lower  potential  and 
thus  constitute  an  electric  current.  They  are  confined  in 
their  motion  entirely  to  the  body  of  the  wire.  The  attrac- 
tion of  the  metallic  molecules  for  the  electrons  is  so  great 
that  the  electrons  are  forced  to  remain  in  the  material 
and  cannot  leave  it.  In  their  progress  down  the  wire,  they 
collide  with  the  metallic  molecules  and  impart  some  of  their 
acquired  energy  to  the  latter.  This  results  in  a  heating 
of  the  conductor  and  a  friction  to  the  motion  of  the  elec- 
trons which  we  know  as  the  resistance  that  the  conductor 
offers  to  a  flow  of  current.  This  resistance  for  a  given  con- 
ductor, we  have  seen,  is  a  constant  so  long  as  the  tempera- 
ture remains  fixed.  It  is  true  that  the  resistivity  of  the 
material  depends  upon  the  temperature  at  which  it  is  used, 

391 


392      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

but  temperature  changes  when  they  occur  are  comparatively 
slow,  so  that  the  flow  of  the  current  at  any  time  can  readily 
be  computed  by  using  the  resistances  of  the  several  parts 
of  the  circuit  at  that  particular  instant. 

Electric  circuits  are  generally  constructed  of  metal  wires, 
so  that  the  above  type  of  conductor  is  most  important  from 
a  practical  standpoint.  There  are,  however,  a  large  num- 
ber of  devices  of  various  sorts  which  utilize  conduction  of 
a  different  nature.  Such  conduction  is  not  in  accordance 
with  Ohm's  law.  This  law  and  the  other  laws  of  circuits 
cannot  be  applied  in  such  exceptions  without  several  modifi- 
cations. It  is  therefore  necessary  that  we  make  special 
study  of  apparatus  to  which  Ohm's  law  does  not  apply. 

107.  Non-Metallic  Conduction  of  Electricity.  One  form 
of  non-metallic  conduction  we  have  studied  under  the 
heading  of  electrolytic  conduction.  This  we  have  seen  to 
consist  of  the  movement  of  charged  ions  through  the  body 
of  a  liquid.  Conduction  through  an  electrolyte  is  accom- 
panied by  a  movement  of  the  molecules  themselves,  and 
hence  is  in  distinction  to  conduction  through  a  metal  where 
the  molecules  of  the  material  itself  are  fixed  in  position. 

Another  type  of  non-metallic  conduction  which  is  of  great 
importance  is  the  conduction  through  gases.  The  laws 
which  govern  conduction  through  gases  are  at  the  basis 
of  the  construction  of  mercury-arc  rectifiers,  arc  lights, 
some  types  of  lightning  arresters,  X-ray  tubes  and  a  multi- 
tude of  other  devices.  They  also  govern  the  corona  loss  from 
transmission  lines  and  much  of  the  behavior  of  insulations. 

In  addition  to  gaseous  conduction  proper,  there  is  a  form 
of  conduction  depending  upon  thermionic  emission.  This 
is  at  the  basis  of  the  thermionic  tube,  which,  not  to  men- 
tion its  use  in  radio  practice  and  other  places,  has  an  ex- 
tensive use  in  telephony  as  a  repeater.  There  are  also 
in  use  several  devices  which  do  not  obey  Ohm's  law,  the 
action  of  which  cannot  be  explained  as  simply  electrolytic 
or  gaseous  in  nature  Among  these  phenomena  may  be 


CONDUCTION  THROUGH  GASES  393 

mentioned  the  conduction  between  a  metal  and  a  crystal 
surface,  where  the  crystal  is  conducting.  This  is  the  crystal 
rectifier  used  in  radio  telegraphy.  However,  this  chapter 
will  be  confined  to  thermionic  conduction  and  conduction 
through  gases. 

108.  Thermionic  Conduction.  In  the  early  days  of  the 
electric  light,  it  was  discovered  by  Edison  that  if  a  cold 
metal  plate  was  placed  in  an  electric  light  bulb,  a  current 
could  be  passed  between  the  plate  and  the  heated  filament 
if  a  potential  was  applied  between  them.  In  fact,  he  found 
that  while  a  current  could  be  made  to  flow  through  the 
vacuum  in  the  direction  from  the  plate  to  the  filament,  no 
current  whatever  would  flow  in  the  opposite  direction,  re- 
gardless of  how  great  potential  was  applied,  up  to  the  break- 
down point.  This  device,  then,  was  a  rectifier  which  would 
change  alternating  into  direct  current.  Its  current  capac- 
ity, that  is,  the  amount  of  current  which  could  be  passed 
through  it,  was  very  small,  and  hence  it  did  not  find  a  use 
as  a  power  rectifier.  Fleming,  however,  discovered  that 
the  rectifying  properties  of  the  device  were  still  present  at 
very  high  frequencies.  He  accordingly  made  use  of  this 
device  as  a  thermionic  valve  to  rectify  the  incoming  high- 
frequency  current  of  a  radio-telegraph  signal,  and  thereby 
to  change  it  into  direct  current  so  that  it  could  affect  a 
receiving  instrument.  This  device  then  "became  the  most 
sensitive  detector  for  radio  telegraphy  then  available. 

The  current  between  the  filament  and  the  plate  is  evi- 
dently not  conducted  in  any  ordinary  manner.  Experiment 
shows  that  the  conduction  of  the  device  is  governed  by 
unique  laws.  If  the  potential  between  filament  and  plate 
of  a  tube  constructed  and  connected  as  shown  in  Fig. 
194  is  varied,  the  current,  measured  by  an  ammeter  at  A, 
will  vary  in  accordance  with  a  characteristic  curve  such 
as  is  shown  in  Fig.  195. 

The  device  evidently  does  not  obey  Ohm's  law.  If  it 
did,  its  characteristic  curve  would  be  a  straight  line.  It 


394      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Plate  voltage  Voltmeter 
Plate  current  Ammeter 


Filament  Rheostat 

FIG.  194.  When  the  filament  is  heated  and  an  electric  potential  is  ap- 
plied between  the  filament  and  the  cold  plate,  electrons  pass  from  the 
filament  to  the  plate. 


Plate  Voltage 

FIG.  195.  The  curve  shows  the  relation  existing  between  plate  potential 
and  plate  current  if  the  filament  temperature  of  Fig.  194  is  main- 
tained constant. 


CONDUCTION   THROUGH  GASES 


395 


does  not  have  a  constant  resistance,  but  its  resistance  varies 
according  to  the  current  passing  through  the  device.  A 
study  of  this  characteristic  curve  shows  several  things.  In 
the  first  place,  there  is  a  current  which  flows  when  the  plate 
is  positive,  but  no  current  when  it  is  negative.  Second, 
the  current  rises  slowly  at  first  and  then  more  rapidly  as 
the  voltage  is  increased,  and  finally  reaches  a  maximum 
or  saturation  value  which  is  not  increased  no  matter  how 
high  the  potential  is  made.  The  curve  shown  in  Fig.  195  is 
for  one  given  value  of  filament  current.  If  the  filament 
current  is  adjusted  to  several  successive  values,  a  series  of 
curves  can  be  obtained  as  shown  in  Fig.  196,  each  curve 


Plate  Voltage 

FIG.  196.  The  curves  show  the  relations  between  plate  current  and 
plate  voltage  for  different  temperatures;  the  higher  the  temperature, 
the  higher  the  saturation  plate  current. 

being  for  a  different  value  of  filament  temperature.  It 
will  be  noted  that  the  curves  for  different  values  of  filament 
current  coincide  for  low  values  of  plate  voltage,  but  that 
the  saturation  value  of  current  reached  in  each  case  is  differ- 
ent, and  is  higher  the  higher  the  filament  current  and  hence 
the  filament  temperature. 

109.  Richardson's  Law  of  Thermionic  Emission.    The 
behavior  of  the  above  device  may  be  explained  as  follows. 


396      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

As  the  metal  of  the  filament  is  heated,  the  free  electrons, 
which  are  in  the  body  of  the  metal,  move  faster  by  reason 
of  their  heat  velocity.  In  fact,  if  the  temperature  is  suffi- 
ciently increased,  some  of  the  electrons  will  move  so  rapidly 
that  they  will  escape  from  the  body  of  the  metal  and  fly  out 
into  space.  This  effect  is  exactly  analogous  to  the  evapora- 
tion of  a  liquid.  When  a  liquid  is  heated,  some  of  the 
molecules  by  reason  of  their  heat  motion  acquire  enough 
speed  to  be  able  to  escape  from  the  surface.  In  exactly  the 
same  manner,  we  may  evaporate  electrons  from  an  incan- 
descent filament. 

If  the  filament  is  made  negative  with  respect  to  the  cold 
plate,  these  electrons  are  repelled  by  the  filament  and  at- 
tracted by  the  plate.  Some  of  them  accordingly  move 
across  to  the  plate.  This  constitutes  an  electric  current 
which  is  measured  by  the  ammeter.  The  direction 
of  this  current  we  take  in  accordance  with  convention  to 
be  in  the  opposite  direction  to  that  in  which  the  electrons 
actually  move.  This,  as  we  have  seen,  is  because  of  the 
fact  that  due  to  a  nomenclature  which  was  adopted  before 
the  electron  theory  was  known,  the  charge  on  an  electron 
is  found  to  be  negative.  Of  course  it  makes  no  difference 
in  which  direction  we  draw  the  arrow  indicating  the  direc- 
tion of  current  provided  we  are  consistent.  The  motion 
of  the  electron,  however,  is  always  from  the  hot  filament 
to  the  cold  plate,  and  never  in  the  other  direction;  for  the 
cold  plate  does  not  give  out  any  electrons  by  which  a  current 
could  be  set  up  in  the  opposite  direction. 

When  the  plate  voltage  is  small,  only  a  few  of  the  elec- 
trons evaporated  from  the  filament  move  over  to  the  plate, 
the  remainder  returning  to  the  filament.  As  the  voltage 
is  increased,  however,  larger  and  larger  numbers  of  the 
electrons  move  over  to  the  plate;  that  is,  the  current  in- 
creases. If  the  potential  between  filament  and  plate  is 
sufficiently  increased,  all  of  the  electrons  will  pass  over 
to  the  plate  as  fast  as  they  are  evaporated  from  the  filament. 


CONDUCTION  THROUGH  GASES  397 

Then  no  matter  how  much  the  potential  increases  beyond 
this,  the  current  cannot  increase,  for  saturation  has  been 
reached.  This  accounts  for  the  saturation  value  of  current 
which  is  shown  on  the  curve  of  Figure  195,  and  which  is 
the  current  which  cannot  be  exceeded  in  the  device  at 
the  given  value  of  filament  current.  If  the  filament  cur- 
rent is  increased,  the  temperature  is,  of  course,  increased 
and  hence  a  larger  number  of  electrons  per  second  is  evap- 
orated. Then  the  maximum  current  which  can  be  passed 
through  the  device  is  also  increased,  or  the  saturation  current 
becomes  greater.  This  gives  rise  to  the  series  of  curves 
shown  in  Figure  196. 

The  evaporation  of  electrons  described  above  is  known 
as  thermionic  emission,  and  a  device  which  utilizes  such 
an  effect  is  a  thermionic  valve.  It  is  not  gaseous  conduction, 
for  there  is  no  gas  which  plays  a  part  in  the  device.  In 
fact,  the  more  completely  the  bulb  is  exhausted  of  air, 
the  better  the  device  works.  Thermionic  tubes  are  accord- 
ingly freed  as  far  as  possible  from  gas ;  that  is,  they  operate 
in  a  hard  vacuum.  The  conduction  is  hence  across  an 
evacuated  space  by  reason  of  electrons  which  are  evapor- 
ated from  the  filament  and  which  fly  from  the  filament  to 
the  plate  through  empty  space. 

The  law  which  governs  the  number  of  electrons  evapor- 
ated from  a  filament  was  discovered  by  Richardson.  It  is  of 
exactly  the  same  form  as  the  law  that  governs  evaporation 
from  a  liquid  and  is  therefore  exponential.  The  number 
of  electrons  evaporated  can  be  measured  by  determining 
the  saturation  current  of  the  device.  In  this  form,  Richard- 
son's law  may  be  expressed  thus: 

i-ar*«~T'  (1) 

where 

is  is  the  saturation  current  in  amperes, 
T  is  the  temperature  of  the  filament  in  centigrade  degrees 
absolute, 


398      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

a  is  a  constant  depending  upon  the  length,  diameter  and 

material  of  the  filament, 
K  is  a  constant  depending  upon  the  material  of  the 

filament  only. 

For  drawn  tungsten  wire,  the  value  of  K  is  almost  exactly 
52,500.  For  tungsten  also  we  find  that  a  has  the  value 
approximately  23.6  X  106  times  the  surface  of  the  filament 
in  square  centimeters.  In  utilizing  the  formula,  we  must  be 
careful  to  note  that  it  applies  only  to  the  portion  of  the 
filament  which  is  actually  at  the  temperature  T,  and  that 
the  ends  which  are  cooled  by  the  support  will  emit  only  to 
a  lesser  degree. 

Example  1.  As  an  example,  if  we  have  a  filament  5  centi- 
meters long  and  of  0.05  millimeter  diameter,  of  tungsten,  which 
is  at  a  temperature  of  2200  degrees  absolute,  the  saturation 
current,  that  is,  the  maximum  current  which  can  be  passed 
through  a  valve  with  the  filament  at  this  temperature,  will  be 

is  =  23.6  X  106  X  5  X  TT  X  0.005  V2200e-52500/220° 
=  0.004  ampere. 

According  to  Richardson's  formula,  the  filament  will  emit 
somewhat  at  any  temperature.  At  ordinary  temperatures, 
however,  the  emission  is  very  slight.  This  can  be  checked  by 
computing  the  current  from  the  above  filament  at  a  temperature 
of  say  100  degrees  centigrade,  that  is,  373  degrees  absolute. 
The  current  will  be 

is  =  23.6  X  106  X  5  X  TT  X  0.005  V373  f-52500/373 
=  4  X  10-53ampere 

which,  it  will  be  seen,  is  an  exceedingly  small  current  and  quite 
negligible. 

The  electrons  are  speeded  up  by  the  action  of  the  plate 
voltage  and  impinge  upon  the  plate  at  high  velocity.  This 
naturally  heats  the  plate,  and  if  a  sufficient  current  passes 
through  the  device  at  a  high  voltage  drop,  the  plate  will 
become  very  hot.  Of  course,  it  cannot  be  allowed  to  become 
so  hot  that  it  will  also  emit  electrons  to  any  great  extent,  for 


CONDUCTION   THROUGH  GASES  399 

in  that  case  the  device  will  cease  to  rectify.  The  loss  in 
the  plate  circuit  is,  of  course,  equal  to  the  current  multiplied 
by  the  drop  of  potential  in  the  tube.  This  loss  in  watts 
all  appears  as  heat  in  the  plate.  In  addition  there  is  an 
expenditure  of  energy  required  to  keep  the  filament  hot. 

The  thermionic  rectifier  is  beginning  to  be  of  importance 
in  other  fields  than  that  of  radio  communication.  In  the 
next  decade  it  will  undoubtedly  be  used  to  considerable  ex- 
tent even  in  power  work.  Its  current-carrying  capacity  is 
somewhat  strictly  limited,  but  the  voltages  it  can  handle  are 
high.  Even  a  current  of  1  ampere  at  a  potential  of  100,000 
volts  represents  a  power  of  100  kilowatts,  and  a  tube  which 
can  rectify  this  amount  is  not  a  toy.  In  connection  with 
controls  by  means  of  grids  or  magnetic  fields  it  will  be 
used  for  circuit  interruption  and  similar  service.* 

Prob.  1-12.  A  thermionic  valve  is  constructed  with  a  fila- 
ment 2  inches  long  and  0.005  inch  in  diameter.  At  a  filament 
current  of  1.1  amperes  this  filament  is  at  a  temperature  of 
2100  degrees  absolute.  A  length  of  f  inch  on  each  end  of 
the  filament  is  cooled  by  the  support  and  may  be  considered 
not  to  emit.  A  voltage  of  1000  volts  applied  between  the  fila- 
ment and  the  plate  causes  the  saturation  current  to  flow.  Com- 
pute the  value  of  this  saturation  current  and  the  watts  trans- 
formed into  heat  in  the  plate. 

Prob.  2-12.  After  the  device  in  Prob.  1-12  has  been  operated 
steadily  for  a  time,  the  plate  is  observed  'to  come  to  a  tem- 
perature of  400  degrees  centigrade.  Assuming  that  the  heat 
dissipation  from  the  plate  is  proportional  to  the  fourth  power 
of  the  absolute  temperature,  compute  the  temperature  to 
which  the  plate  will  come  if  the  filament  temperature  is  in- 
creased to  2200  degrees  absolute,  the  voltage  across  the  tube 
remaining  unchanged  and  being  sufficient  to  produce  satu- 
ration. 

Prob.  3-12.  If  the  mean  temperature  coefficient  of  tung- 
sten is  taken  as  0.006,  what  filament  potential  will  be  necessary 
to  produce  1.2  amperes  in  the  filament  of  Prob.  1-12?  Assume 
800°  C.  as  the  temperature  of  £  in.  at  each  end  of  filament. 

*  See  prediction  in  paper  by  A.  W.  Hull,  Journal  A.  I.  E.  E.,  1921. 


400      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Prob.  4-12.  Will  the  current  be  the  same  at  all  points 
along  the  filament  of  Prob.  1-12?  If  not,  by  what  percent- 
age will  it  vary  from  a  mean 
value,  assuming  the  filament  to 
emit  uniformly?  What  effect 
will  this  have  on  the  behavior 
of  the  valve? 


110.  Thermionic  Amplifiers 
and  Oscillators.  De  Forest 
placed  a  third  electrode  in  a 
thermionic  tube  and  obtained 
remarkable  results.  This  third 
electrode  is  in  the  form  of  a 
grid  located  between  the 
filament  and  the  plate  as 
shown  in  Fig.  197.  The  de- 
velopment of  this  idea  by  a 
large  number  of  investigators 
has  made  possible  present-day 
long-distance  telephony.  The  repeaters  used  on  the  long 
lines  of  the  American  Telephone  and  Telegraph  Company 
are  now  nearly  all  of  this  type.  One  of  the  tubes  is  shown 
in  Fig.  198. 

When  a  voltage  is  established  between  filament  and  plate 
in  a  three-electrode  tube,  the  electrons  in  their  flight  leave 
the  hot  electrode,  pass  between  the  meshes  of  the  grid  and 
finally  arrive  at  the  plate.  A  small  change  of  potential 
between  the  grid  and  filament  will  then  greatly  affect  the 
electrons'  flow.  This  is  shown  in  the  curves  of  Fig.  199, 
which  give  grid  characteristics  for  a  common  type  of  ampli- 
fier tube  plotted  for  different  values  of  plate  voltage.  It 
will  be  noted  that  a  small  change  in  grid  voltage  may  be 
made  to  cause  a  large  change  in  plate  current.  If  the  grid 
is  negative,  as  may  be  made  the  case  by  connecting  in  a  small 
grid  battery  as  shown  in  Fig.  200,  there  will  be  no  grid  cur- 
rent whatever,  for  the  grid  will  at  all  times  repel  the  electrons. 


FIG.  197.  A  thermionic  tube 
with  a  grid  introduced  between 
the  filament  and  the  cold  plate. 


CONDUCTION   THROUGH  GASES 


401 


The  input  to  the  grid  circuit  can  hence  be  made  very  small. 
We  can  thus  control  a  large  amount  of  output  energy  by  an 


.2  5 
£  £ 

^     cc 


expenditure  of  practically  no  energy  at  all.  Any  device 
which  in  this  way  enables  a  small  amount  of  input  power  to 
vary  a  large  amount  of  output  power  is  an  amplifier. 

It  will  be  noted  in  Fig.  199  that  for  a  certain  range  of 
grid  potential,  the  characteristic  is  really  a  straight  line. 


402      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

If  we  operate  the  tube  over  this  range  only,  the  plate  cur- 
rent will  vary  in  linear  relation  to  the  grid  potential.  This 
is  called  distortionless  amplification. 

Amplifiers  are  used  for  very  many  purposes.     Principal 


FIG.    199.     The  relation  between  grid  potential  and  plate  current. 
From  Morecroft's,  "Principles  of  Radio  Communication". 

among  these  is  to  amplify  telephone  currents.  A  telephone 
current  is  usually  a  current  which  is  varying  in  accordance 
with  the  speech  of  the  user.  It  is  an  alternating  current  of 
rapidly  varying  frequency  and  wave  form.  Cause  this 
varying  current  to  impress  an  exactly  similar  varying  volt- 


CONDUCTION   THROUGH  GASES 


403 


age  upon  the  grid  of  an  amplifying  tube  adjusted  to  the 
straight  portion  of  its  characteristic.  The  result  is  that 
the  plate  current  will  vary  in  just  the  way  that  the  input 
does  but  its  variation  will  be  of  much  greater  amplitude. 
A  weak  current  may  thus  be  amplified  almost  indefinitely. 


FIG.  200.     The  thermionic  tube  connected  as  an  amplifier. 

The  action  of  the  oscillator  follows  from  the  above.  In 
fact,  any  amplifier  of  energy  may  be  thus  used  as  an  oscil- 
lator, by  deflecting  part  of  its  output  back  on  its  input  cir- 
cuit. Suppose,  for  example,  that  we  have  a  thermionic  tube 
with  an  input  of  0.01  watt  at  1000  cycles  and  an  output  of 
10  watts.  We  may  take  a  small  part,  about  0.01  watt,  of  this 
large  output,  and  convert  it  by  means  of  transformers  so 
that  it  becomes  input  to  the  tube.  We  may  then  dispense 
with  any  other  input,  and  the  tube  will  continue  to  operate 
or  oscillate.  In  order  that  the  frequency  may  remain  at 
1000  cycles,  it  will  be  necessary  to  supply  an  oscillating  cir- 
cuit consisting  of  an  inductance  and  a  capacity,  the  natural 
frequency  of  which  is  1000  cycles.  One  circuit  by  which 
this  may  be  accomplished  is  shown  in  Fig.  201. 

111.  X-Ray  Tubes.  An  electron  is  a  negative  charge  of 
electricity.  When  situated  between  two  plates  which  are  of 
different  polarities,  it  is  attracted  by  one  and  repelled  by  the 
other.  Accordingly  it  is  set  in  motion;  that  is,  it  receives 
acceleration.  The  acceleration  which  it  receives  is  pro- 


404      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


portional  to  the  force  on  the  electron,  and  this  is  equal 
to  the  charge  of  an  electron  multiplied  by  the  potential 
gradient  at  the  point  in  question.  By  potential  gradient 
we  mean  the  volts  per  centimeter  which  exist  at  the  point 

in  question.  It  is  meas- 
ured in  abvolts  per  centi- 
meter, if  we  are  working 
in  the  c.g.s.  system.  If 
the  electric  field  is  uni- 
i  form,  the  potential  grad- 

/JvWV-  ient  is  equal  to  the  volt- 

age between  the  plates 
divided  by  the  distance 
between  them.  That  is, 
if  1000  volts  exist  be- 
tween plates  which  are 
three  centimeters  apart, 
we  have  a  potential 
gradient  between  them 

of  333  volts  per  centimeter.  If  the  field  is  not  uniform,  the 
gradient  varies  from  point  to  point. 

When  we  have,  in  an  evacuated  device,  a  heated  filament 
and  a  cold  plate  between  which  electrons  pass,  the  speed 
which  the  electrons  attain  may  be  computed  in  the  following 
manner.  In  passing  between  filament  and  plate,  their 
change  of  potential  energy  is,  of  course,  the  integral  of  force 
times  the  distance  between  the  plate  and  the  filament. 
This  will  be  equal  to  the  kinetic  energy  which  they  acquire. 
This  gives  rise  to  the  equation 


FIG.   201. 


The  thermionic  tube  as  an 
oscillator. 


/mv2 
fdx  =  -3- 


(2) 


If  X  is  the  potential  gradient  and  q  is  the  charge  on  the 
electron,  then 

/  =  Xq  (3) 


CONDUCTION  THROUGH  GASES  405 

and 

ffdx  =   CqXdx. 

But  JXdx  is  simply  the  voltage  between  filament  and  plate. 
Accordingly 

Eq  =  '^~-  (4) 

or  

v  =  \/2E<L  (5) 

This  equation  gives  us  a  means  of  finding  the  velocity  which 
an  electron  will  acquire  in  falling  through  a  given  difference 
of  potential.  The  quantity 

I'  .:;,,   <6> 

which  occurs  in  this  equation,  is  the  ratio  of  the  charge  to  the 
mass  of  an  electron.  This  constant  is  of  fundamental  im- 
portance and  has  been  determined  by  a  large  number  of 
different  methods  of  measurement.  Its  value  when  q  is 
in  electromagnetic  units  and  m  in  grams  is  approximately 

1,77  X  107. 

Let  us  compute  an  example.     Suppose  that  there  is  a  po- 
tential of  1000  volts  between  filament  and  plate;  that  is, 

E  =  1011  abvolts. 
Substituting  in  our  equation  above,  we  obtain  for  the  velocity 


v  =  V2  X  1011  X  1.77  X  107 

=  1.9  X  109  centimeters  per  second. 

This  is  indeed  a  very  high  velocity.     It  is  more  striking  if 
we  convert  it  into  the  value 

v  =  12,000  miles  per  second. 

It  is  naturally  to  be  expected  that  an  electron  moving  with 
this  velocity  will  produce  a  peculiar  effect  when  it  suddenly 


406      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


strikes  the  hard  surface  of  the  metal.  This  is  in  fact  the 
case,  and  it  has  been  found  that  when  electrons  of  very  high 
velocity,  corresponding  to  a  potential 
drop  of  10,000  to  50,000  volts,  impinge 
upon  the  metal  plate,  the  plate  gives 
out  a  form  of  light  of  very  short  wave 
length  which  penetrates  even  very 
dense,  opaque  objects.  These  rays 
are  known  as  X-rays  and  are  of  very 
great  service  in  medical  practice  and 
in  various  other  connections. 

The  form  of  X-ray  tube  employing 
a  heated  filament  and  a  cold  plate  or 
target  is  known  as  the  Coolidge  X-ray 
tube.  It  is  arranged  as  shown  in 
Fig.  202. 

In  order  to  intensify  the  effect  of 
such  a  tube,  an  arrangement  is  used 
whereby  the  electrons  are  focused  on 
a  small  spot  on  the  target.  This  is 
accomplished  by  means  of  a  shield 
placed  around  the  filament  and  con- 
nected to  the  filament.  The  shield  as 
well  as  the  filament  repels  the  elec- 
trons when  they  are  first  emitted  and  when  they  are 
moving  at  low  velocity.  By  properly  shaping  this  shield, 
they  are  so  directed  during  the  initial  stages  of  their  flight 
that  they  are  aimed  toward  a  certain  spot  on  the  target. 
After  they  have  acquired  a  high  speed,  it  is,  of  course,  con- 
siderably more  difficult  to  deflect  them.  The  direction 
which  they  take  is  therefore  determined  almost  entirely 
during  the  first  part  of  their  flight. 

A  Coolidge  X-ray  tube  may  employ  potentials  as  high  as 
100,000  volts.  Either  an  alternating  or  a  direct  potential 
may  be  used.  When  an  alternating  potential  is  used,  the 
tube  itself  does  the  necessary  rectifying. 


FIG.  202.    A  Coolidge 
X-ray  tube. 


CONDUCTION  THROUGH  GASES  407 

The  electrons  in  their  motion  are,  of  course,  invisible. 
So,  for  that  matter,  are  the  X-rays.  However,  when  elec- 
trons or  ions  impinge  upon  the  glass  of  the  tube,  they  cause 
it  to  fluoresce.*  When  an  X-ray  tube  is  in  operation, 
a  hemisphere  of  the  bulb  is  accordingly  fluorescent  with  a 
greenish  light.  This  hemisphere  is  roughly  bounded  by 
the  plane  of  the  target  extended.  The  X-rays,  of  course, 
are  emitted  from  the  target  in  all  directions,  and  not  focused 
as  is  the  stream  of  electrons. 

A  Coolidge  X-ray  tube  operates  in  a  hard  vacuum; 
that  is,  all  of  the  gas,  so  far  as  possible,  is  exhausted  from 
the  bulb.  The  presence  of  even  a  very  minute  amount 
of  gas  will  prevent  the  proper  operation  of  the  device. 

A  molecule  consists,  as  we  know,  of  a  positive  nucleus 
surrounded  by  a  number  of  electrons  more  or  less  securely 
attached  to  the  molecule.  When  an  electron  at  high  ve- 
locity encounters  such  a  system,  it  sometimes  knocks  off 
one  or  more  electrons.  This  process,  we  have  seen  in  Chap- 
ter V,  is  called  ionization.  The  electrons  thus  produced, 
together  with  the  original  electrons,  proceed  together  to 
the  positive  plate.  The  remainder  of  the  molecule,  now 
lacking  an  electron,  and  therefore  positively  charged,  is 
propelled  in  the  opposite  direction  and  hits  the  filament. 

The  movement  of  these  extra  electrons  and  positive  ions 
is  also  a  movement  of  electricity  and  constitutes  an  electric 
current.  Hence  one  effect  produced  when  there  is  ioniza- 
tion in  a  thermionic  device  is  an  increase  in  current  through 
the  tube. 

There  is  another  important  effect,  however.  The  positive 
ions  are  large  and  comparatively  heavy.  When  they  im- 
pinge upon  a  metal  surface,  it  is  possible  to  cause  them 
to  emit  electrons  by  the  process  of  knocking  them  bodily 
out  of  the  surface  of  the  metal.  It  is  thus  possible  to  get  con- 
duction through  a  tube  without  utilizing  a  heated  filament. 

*  A  body  is  said  to  fluoresce  when  it  throws  out  light  waves  of 
wave  lengths  different  from  any  which  it  is  receiving. 


408      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

In  fact,  it  was  in  this  manner  that  X-ray  tubes  were 
first  constructed.  A  small  amount  of  gas  was  allowed  to 
remain  in  the  device.  In  any  gas  there  are  always  a  num- 
ber of  free  electrons.  The  number  may  be  as  high  as  1,000 
per  cubic  centimeter  for  air  in  a  normal  state.  This  is,  of 
course,  very  small  when  compared  with  the  total  number  of 
molecules  present.  When  a  potential  is  applied  between 
two  cold  electrodes  in  a  tube  containing  a  small  amount 
of  gas,  these  free  electrons  are  speeded  up  and  fly  toward 
the  positive  electrode.  In  their  path  they  encounter  mole- 
cules and  ionize  them.  The  positive  ions  thus  produced 
impinge  upon  the  negative  electrode  and  cause  it  to  emit 
electrons.  These  in  their  turn  proceed  on  their  way  and 
produce  more  ions.  Thus  the  process  is  continuous. 

The  electrons  leaving  the  surface  of  the  negative  elec- 
trode or  cathode  fly  off  in  a  direction  almost  perpendicular 
to  the  surface.  By  making  the  cathode  a  section  of  a 
spherical  surface  with  the  target  at  the  center  of  the  sphere, 
the  electrons  may  be  concentrated,  and  such  of  them  as  do 
not  encounter  molecules  on  their  way  will  impinge  upon 
the  target  within  a  comparatively  small  area.  This  is  the 
older  type  of  X-ray  tube. 

An  X-ray  tube  of  this  sort  is  very  sensitive  to  gas  pres- 
sure. -If  there  is  too  little  gas  present,  no  current  at  all  can 
be  passed,  for  in  order  that  it  shall  be  possible  to  pass  a 
current,  the  process  of  ionization  must  be  cumulative;  that 
is,  more  ions  must  be  formed  by  collision  than  are  lost  by 
absorption  into  the  electrode.  On  the  other  hand,  if  the 
gas  pressure  is  too  high,  it  will  be  possible  to  have  a  current 
but  nearly  all  of  the  electrons  which  leave  the  cathode  will 
hit  molecules  before  they  arrive  at  the  target.  There  are 
simply  too  many  molecules  in  the  way.  Another  way  of 
saying  the  same  thing  is  that  the  mean  free  path  of  the  elec- 
trons in  the  gas  must  be  of  about  the  same  order  of  magni- 
tude as  the  distance  between  the  cathode  and  the  target. 
By  the  mean  free  path  is  meant  the  average  distance  which 


CONDUCTION  THROUGH  GASES  409 

the  electrons  will  fly  in  a  straight  line  before  encountering 
molecules.  If  the  gas  pressure  is  high  and  there  is  a  large  num- 
ber of  molecules  present,  the  mean  free  path  will  be  short. 
In  such  a  case  almost  all  the  electrons  leaving  the  cathode 
will  encounter  molecules  before  they  have  proceeded  far. 
Very  few  of  them  indeed  will  fly  straight  from  cathode  to 
anode  and  impinge  upon  the  target.  It  is  this  latter  ac- 
tion which  we  have  seen  to  be  necessary  for  the  proper  pro- 
duction of  X-rays.  The  pressure  in  a  gaseous  X-ray  tube 
must  hence  be  very  carefully  adjusted  to  its  proper  value. 
This  critical  adjustment  is  very  difficult  to  make  correctly, 
and  this  fact  has  rendered  the  gaseous  type  of  X-ray  tube 
almost  obsolete. 

Prob.  5-12.  The  mass  of  an  electron  is  approximately 
9  X  10~28  gram.  If  we  have  an  X-ray  tube  operating  with  a 
potential  of  10,000  volts  between  filament  and  plate,  what 
energy  will  be  possessed  by  each  electron  when  it  arrives  at 
the  plate?  How  fast  will  it  be  moving,  if  we  assume  the  mass 
to  be  constant?  If  a  current  of  100  milliamperes  is  being  passed 
through  the  tube,  how  much  heat  is  being  produced  at  the  target? 
Assuming  the  target  to  be  of  copper  and  weighing  250  grams, 
how  fast  will  its  temperature  rise  when  the  tube  is  first  started? 

Prob.  6-12.  How  many  electrons  per  second  are  passing 
across  between  filament  and  plate  in  the  tube  of  Prob.  5-12? 

112.  The  Gaseous  Discharge  Tube.  In  a  gaseous  X-ray 
tube,  the  mean  free  path  of  an  electron  in  the  residual  gas  is  of 
about  the  order  of  magnitude  of  the  dimensions  of  the  tube. 
This  means  that  of  the  electrons  which  pass  between  fila- 
ment and  target,  only  a  fair  percentage  collide  with  mole- 
cules on  the  way  and  ionize.'  The  potential  across  the  tube 
necessary  to  pass  even  a  very  small  current  is  accordingly 
correspondingly  high. 

If  we  let  more  gas  into  a  discharge  tube  of  this  sort,  the 
mean  free  path  of  the  electrons  is  decreased;  that  is,  an  electron 
will  travel  much  less  distance  on  the  average  before  encoun- 
tering a  molecule  and  ionizing  it.  It  is  therefore  much  easier 


410      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


FIG.  202a.  The  tungar  ap- 
paratus for  rectifying  small 
alternating  currents.  Gen- 
eral Electric  Co. 


to  pass  a  current  through  the  tube,  because  the  potential 
drop  in  the  tube  for  a  given  current  will  be  less.  In  fact, 
if  a  sufficient  amount  of  gas  is  present,  considerable  current 

may  be  passed  through  the  tube 
with  a  drop  of  only  two  or  three 
hundred  volts.  The  current 
which  can  be  passed  is  therefore 
determined  only  by  the  capacity 
of  the  tube,  that  is,  by  the 
amount  of  loss  that  it  can  stand 
without  becoming  too  hot.  When 
the  pressure  is 
about  one  milli- 
meter of  mer- 
cury, the  tube 
then  containing 
gas  of  an  a- 
mount  approx- 
imately one-seven-hundred-and-sixtieth  of 
the  amount  which  would  be  present  at  at- 
mospheric pressure,  conduction  takes  place 
very  freely.  There  are  so  many  mole- 
cules present  that  an  electron  does  not 
proceed  far  on  its  way  before  it  encounters 
one  of  them  and  ionizes  it.  The  mean 
free  path  under  these  conditions  for  an 
electron  in  hydrogen  is  about  0.7  milli- 
meter. 

When  a  small  amount  of  gas  is  admitted 
to  a  thermionic  tube,  the  current  that  it 
can  carry  is  greatly  increased.  Such  a  tube 
may  be  used  as  a  rectifier  for  charging  stor- 
age batteries  from  an  alternating  potential  source.  The 
"tungar,"  a  tube  of  this  type,  shown  in  Fig.  202a  and  202&, 
can  carry  several  amperes  with  a  drop  of  only  a  few  volts. 
It  cannot,  however,  rectify  high  potentials. 


FIG.  202b.  The 
tungar,  a  tube 
which  operates 
as  a  rectifier. 
General  Electric 
Co. 


CONDUCTION   THROUGH  GASES  411 

Gaseous  conduction  is  accompanied  by  a  brilliant  glow  in 
the  gas.  Every  time  a  molecule  is  ionized,  it  gives  out  a 
spurt  of  light.  This  light  is  of  a  certain  definite  color  de- 
pendent upon  the  nature  of  the  molecule.  If  the  glow  be- 
tween electrodes  in  such  a  device  is  examined  with  a  spec- 
troscope, it  will  be  found  to  consist  entirely  of  certain  defi- 
nite colors  which  appear  as  lines  in  the  spectroscope.  In 
this  way  gases  can  easily  be  identified  and  studied. 

The  discharge  through  the  tube  is  often  very  beautiful 
and  characteristic  in  its  nature.  It  is  divided  into  several 
different  parts,  and  the  phenomena  which  take  place  within 
the  tube  are  quite  complicated  when  examined  completely. 

Near  the  cathode  is  a  dark  space,  called  the  cathode  dark 
space.  The  reason  for  the  existence  of  a  region  close  to 
the  cathode  where  no  ionization  occurs  is  readily  apparent. 
The  electrons,  as  they  leave  the  cathode,  must  travel  a 
certain  distance  before  they  acquire  sufficient  velocity  to 
enable  them  to  ionize.  The  velocity  which  they  must  at- 
tain depends  somewhat  upon  the  gas  which  is  present,  but 
in  general  they  must  fall  through  a  potential  difference  of 
about  twenty  volts  in  order  to  acquire  the  necessary  speed. 
Also,  after  they  have  acquired  this  speed,  they  will  pro- 
ceed a  certain  further  distance  (on  the  average)  before  they 
hit  a  molecule.  There  is  a  region  near  the  cathode,  there- 
fore, where  the  electrons  are  coming  up  to  speed  and  in 
which  no  glow  is  visible.  Directly  on  the  surface  of  the 
cathode  there  is  a  velvety  light  which  is  probably  caused 
by  the  impact  of  the  positive  ions  as  they  knock  electrons 
out  of  the  surface  of  the  metal.  At  the  end  of  the  cathode 
dark  space,  there  is  a  long  beam  of  light  which  extends 
fully  or  almost  to  the  anode.  This  is  called  the  positive 
column.  By  varying  the  pressure  and  the  current  through 
the  device,  the  form  of  the  discharge  may  be  varied  within 
quite  wide  limits. 

Discharge  tubes  of  this  sort  are  being  used  somewhat  for 
lighting  purposes  at  the  present  time.  Such  a  tube  is  the 
so-called  Moore  light.  There  are  other  forms  which  are 


412      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

just  coining  on  the  market,  and  which  are  made  small  enough 
to  be  screwed  into  a  lamp  socket.  Such  a  light  at  the 
present  time  is  not  highly  efficient,  but  it  may  possibly  be  de- 
veloped considerably  further  and  become  a  very  important 
factor  in  lighting.  It  has  the  advantage  that  the  color 
of  the  light  given  off  may  be  adjusted  almost  at  will  by  prop- 
erly choosing  the  gas  which  is  utilized  in  the  tube  and  by 
varying  the  electrical  conditions  under  which  the  tube  op- 
erates. Another  advantage  is  that  there  is  no  filament  in 
the  light  to  be  burned  out  and  thus  shorten  its  life.  At 
the  present  time  it  is  not  particularly  efficient  in  terms 
of  candlepower  per  watt.  The  radiation  given  off  from  the 
ionized  molecules  can  be  almost  entirely  in  the  visible  spec- 
trum; that  is,  it  can  be  light  accompanied  by  little  heat. 
However,  there  is  of  necessity  a  certain  amount  of  energy 
used  in  heating  the  electrodes,  and  this  energy  is  lost.  If  all 
of  the  energy  could  be  put  into  the  glow  itself,  we  should 
have  a  very  efficient  light  indeed.  Even  all  of  the  energy 
put  into  the  glow  does  not  appear  as  light,  for  some  of  it  is 
used  simply  in  heating  the  gas  in  the  tube. 

It  has  been  noted  that  as  the  pressure  is  increased  in  the 
tube,  the  drop  in  the  tube  decreases.  With  a  hard  vacuum, 
we  can,  of  course,  pass  no  current  through  the  tube  unless 
a  heated  filament  is  present,  for  there  are  no  electrons  to 
conduct  the  current.  When  a  little  gas  is  admitted,  such 
as  in  an  X-ray  tube,  a  small  amount  of  current  may  be 
passed  by  a  very  high  potential  indeed.  Admitting  more 
and  more  gas  lowers  the  potential  drop  by  making  it  easier 
for  the  discharge  to  become  cumulative.  What  is  meant 
by  this  expression  will  be  explained  somewhat  in  detail, 
for  it  is  very  important. 

Suppose  that  there  is  a  discharge  tube  consisting  of  two 
electrodes  in  a  gas,  with  a  certain  potential  between  the 
electrodes  and  a  certain  current  flowing  between  them. 
Suppose  that  an  electron  starts  from  the  cathode.  If  it 
-produces  the  ionization  of  a  molecule  on  the  way,  the  two 
electrons  will  proceed  together  toward  the  anode,  the  positive 


CONDUCTION  THROUGH  GASES  413 

ion  produced  will  "impinge  upon  the  cathode,  and  possibly 
collide  with  several  molecules  on  its  way  to  the  cathode,  and 
upon  striking  the  metal  surface  probably  knock  out  other 
electrons.  The  process  is  then  continued.  If  each  electron 
on  its  way  produces  on  the  average  sufficient  ions  to  start 
another  fresh  electron  from  the  cathode,  then  the  process 
can  proceed.  Ions  are,  however,  lost  in  many  ways.  Some 
of  them  recombine  by  meeting  free  electrons,  and  hence 
becoming  neutral  molecules.  Some  of  them,  by  reason  of 
much  jolting  as  they  pass  through  the  gas,  lose  so  much  en- 
ergy that  they  cannot  knock  electrons  off  the  cathode 
when  they  reach  it.  When  the  discharge  is  cumulative,  we 
mean  that  more  and  more  ionization  tends  to  be  produced  in 
the  body  of  the  gas.  Under  these  conditions,  the  potential 
across  the  device  will  fall  if  the  current  is  held  constant, 
until  only  sufficient  ionization  is  produced  to  carry  the  cur- 
rent. On  the  other  hand,  if  an  insufficient  amount  of  ion- 
ization is  produced  to  carry  the  current,  the  potential  will 
necessarily  rise  until  the  electrons  are  speeded  up  and  suffi- 
cient ionization  in  the  body  of  the  gas  is  maintained. 

In  order  to  produce  cumulative  ionization  we  must  have, 
first,  sufficient  potential  to  bring  electrons  up  to  ionizing 
speed,  and  second,  ample  opportunity  for  these  high-speed 
electrons  to  collide  with  neutral  molecules.  In  fact,  if 
electrodes  are  sufficiently  closely  spaced,  we  may  have 
the  first  of  these  conditions  satisfied  without  the  other. 
Thus  a  short  gap  in  a  gas  may  be  made  to  be  much  harder  to 
break  down  than  a  long  one,  provided  all  the  discharge  paths 
of  the  former  are  short.  The  recognition  of  this  principle 
has  recently  enabled  C.  G.  Smith  to  produce  a  gas-filled  de- 
vice which  can  rectify,  oscillate  and  perform  other  similar 
functions.* 

As  the  pressure  of  the  gas  is  increased  and  the  number  of 
molecules  between  the  electrodes  goes  up,  it  is  easier  to  pro- 
duce cumulative  ionization,  and  accordingly  the  potential 
drop  across  the  device  for  a  given  current  will  go  down. 

*  See  Bush  and  Smith,  "A  New  Rectifier,"  Proc.  I.  R.  E.,  Feb.,  1922. 


414      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

There  is,  however,  a  limit  to  this  effect.  If  the  pressure  of 
the  gas  is  sufficiently  increased,  the  collisions  of  electrons 
with  molecules  will  be  so  frequent  that  they  will  not  have 
time  to  come  up  to  ionizing  velocity  before  impact.  In 
other  words,  an  electron  as  it  flies  between  cathode  and  an- 
ode will  be  continually  jostled  by  the  molecules  and  will  lose 
its  energy.  Between  impacts  with  molecules  it  will  speed 
up,  but  seldom  come  to  so  high  a  velocity  that  upon  striking 
a  molecule  it  can  knock  it  apart.  As  the  pressure  is  in- 
creased beyond  a  certain  critical  value,  the  conditions  for 
ionization  are  therefore  not  so  favorable,  and  the  potential 
across  the  device  will  go  up.  The  gas  pressure  which  will 
give  a  minimum  drop  in  the  tube  in  most  gases  is  one  or 
two  millimeters  with  ordinary  electrode  spacing.  At  higher 
gas  pressures,  the  potential  is  determined  by  somewhat 
different  conditions  than  it  is  at  pressures  below  this  value. 
We  have  seen  that  below  this  critical  value  of  pressure, 
ionization  occurs  smoothly  through  the  body  of  the  gas, 
and  the  potential  across  the  tube  is  regulated  automatically 
to  a  constant  value  dependent  upon  the  current.  To  state 
it  another  way,  the  device  is  stable.  Above  the  critical 
pressure,  however,  we  have  no  longer  stability.  No  current 
at  all  will  be  passed  until  sufficient  potential  is  reached  so 
that  somewhere  in  the  body  of  the  gas  electrons  are  speeded 
up  sufficiently  between  impact  to  produce  ionization.  That 
is,  we  must  produce  sufficient  potential  gradient  at  some 
point  in  the  gas  so  that  the  electrons  in  falling  through  a 
distance  equal  to  their  mean  free  path  will  acquire  an  ion- 
izing velocity.  For  gas  at  atmospheric  pressure,  this  means 
that  a  potential  gradient  of  about  30,000  volts  per  centimeter 
must  be  produced  in  order  to  start  ionization. 

When  ionization  is  started,  however,  it  is  very  rapidly 
cumulative  indeed.  There  is  plenty  of  gas  present  to  ionize, 
and  hence  the  potential  across  the  space  rapidly  goes  down 
and  the  current  goes  up.  The  discharge  is  therefore  sudden 
and  may  be  explosive.  This  is  what  we  know  as  a  spark. 


CONDUCTION  THROUGH  GASES  415 

It  is  usually  accompanied  by  other  effects  which  we  will 
study  in  succeeding  sections. 

Prob.  7-12.  If  it  requires  a  potential  gradient  of  30,000 
volts  per  centimeter  to  break  down  air  at  atmospheric  pressure, 
and  if  we  assume  that  an  electron  speed  of  2  X  108  centimeters 
per  second  is  necessary  in  order  to  ionize,  about  what  is  the 
mean  free  path 'of  an  electron  in  air  at  atmospheric  pressure? 

Prob.  8-12.  Why  is  it,  under  the  conditions  which  are  out- 
lined in  Prob.  7-12,  that  a  spark  discharge  in  air  at  atmospheric 
pressure  is  in  a  narrow  path  between  electrodes,  whereas  a 
glow  discharge  at  low  pressure  fills  the  entire  body  of  the  tube? 

113.  The  Spark.  The  spark,  we  have  seen,  is  a  disruptive 
discharge.  It  has  a  steeply  falling  negative  character- 
istic. This  means  that  the  greater  the  current,  the  less  is 
the  voltage  drop  in  the  spark.  In  fact,  in  air  at  atmos- 
pheric pressure  it  requires  a  very  high  potential  to  produce 
any  appreciable  current  between  separated  electrodes,  but 
when  the  air  breaks  down,  by  sufficient  potential  being  im- 
pressed to  give  ionizing  velocity  to  the  electron,  the  potential 
drop  is  lowered  to  a  comparatively  small  value.  The 
passage  of  a  spark  is  thus  a  sudden  affair,  and  soon  over,  for 
the  large  current  which  can  flow  rapidly  reaches  the  limit 
of  the  source.  If  there  is  much  power  behind  the  spark, 
as  for  instance  when  it  is  over  the  insulator  of  a  high  voltage 
transmission  line,  the  spark  may  be  followed  by  a  destructive 
power  arc  as  we  shall  see. 

In  order  to  produce  a  spark  we  must  stress  the  air  at  some 
point  to  30  kilovolts  per  centimeter.  More  exactly,  this 
value  varies  directly  with  the  density  of  the  air,  and  is  30 
kilovolts  per  centimeter  under  standard  conditions  of  tem- 
perature and  pressure,  25  degrees  centigrade  and  76  centi- 
meters barometer  height.  The  density  5  at  any  other 
temperature  t  and  barometer  height  b  in  centimeters  is  given 
by  the  formula 

3.926 
273  +  t' 


416      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

If  the  potential  is  applied  between  parallel  plates  where  the 
edge  is  protected  in  some  manner,  the  critical  potential  will 
be  reached  when  the  voltage  applied  is  30,000  times  the 
separation  in  centimeters.  When  this  potential  is  reached, 
the  air  will  break  down  suddenly  all  the  way  between  the 
plates  and  a  spark  will  pass. 

With  other  shaped  electrodes  this  simple  rule  will  not, 
however,  apply,  for  the  potential  gradient  will  not  be  uni- 
form, and  it  is  simply  necessary  to  bring  the  air  at  one 
point  up  to  the  critical  gradient. 


FIG.  203.     The  electrostatic  field  between  two  spherical  electrodes. 

An  estimate  of  the  factor  introduced  in  this  manner  may 
be  made  by  mapping  the  electrostatic  field  between  the 
electrodes,  that  is,  by  drawing  the  lines  of  electrostatic 
stress.  In  Fig.  203  this  field  is  shown  for  a  pair  of  spher- 
ical electrodes,  forming  a  sphere  gap.  We  do  not  need 
to  study  the  method  of  drawing  these  fields  accurately. 
Their  approximate  form  may  readily  be  obtained  as  follows. 
The  laws  governing  the  distribution  of  electrostatic  stress 
between  electrodes  are  the  same  as  those  governing  the  distri- 


CONDUCTION   THROUGH  GASES  417 

bution  of  magnetic  flux  in  an  air  gap.  If  we  considered  the 
spherical  electrodes  of  Fig.  203  as  the  iron  portions  of  a 
magnetic  circuit  and  mapped  the  magnetic  field  in  the  gap, 
we  should  obtain  just  the  same  picture. 

The  density  of  the  lines  is  proportional  to  the  potential 
gradient.  The  average  potential  gradient  is  the  total  volt- 
age divided  by  the  separation.  The  ratio  of  the  average 
to  the  maximum  density  of  the  lines  gives  a  factor  to  apply 
to  the  gap  in  obtaining  its  critical  potential.  Thus  if  this 
ratio  is  0.6  and  the  gap  is  two  centimeters  long,  it  will  require 

0.6  X  2  X  30  =  36  kilovolts 

to  break  the  gap  down.  When  this  voltage  across  the  gap 
is  reached,  the  air  close  to  the  electrodes  will  ionize  or  break 
down.  This  layer  of  air  close  to  the  electrodes  then  becomes 
conducting  so  that  the  full  voltage  is  applied  across  the  re- 
mainder of  the  space.  This  raises  the  gradient  on  the  next 
layer  of  air,  which  in  turn  breaks  down,  and  so  on.  Thus 
as  soon  as  the  gradient  in  one  part  of  the  gap  is  reached, 
a  spark  passes  completely  across. 

The  more  the  lines  of  electrostatic  stress  are  crowded 
together,  the  easier  will  a  gap  break  down.  Small  balls 
with  a  given  separation  break  down  at  a  smaller  potential 
than  large  balls. 

The  Standardization  Rules  of  the  American  Institute  of 
Electrical  Engineers  give  the  spark  gap  settings,  for  example, 
for  110,000  volts  break-down  at  25°  C.  and  76  centimeters 
barometer 

for  125. mm.  spheres,  spacing  110  mm; 
"    250  "  "         90      "; 

"   500  "  "         83      "   . 

A  calibrated  spark  gap  may  be  used  for  measuring  very  high 
voltages  by  noting  the  separation  when  the  gap  breaks 
down.  For  a  given  separation  the  easiest  gap  to  break  down 
is  the  needle  gap,  for  the  stress  is  highly  concentrated  at  the 
point  of  the  needle. 


418       PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

All  of  these  matters  pertain  to  a  gradually  applied  voltage. 
Where  a  gap  breaks  down  close  to  the  electrode,  and  then 
spreads  across  as  the  stress  is  redistributed,  it  takes  time 
for  conduction  to  take  place  in  the  ionized  region,  and  hence 
for  the  discharge  to  spread.  For  very  suddenly  applied 
potentials,  therefore,  the  concentration  of  stress  is  relatively 
ineffective.  For  a  sudden  surge  of  voltage  of  very  short 
duration,  all  gaps  of  the  same  separation  will  break  down 
at  the  same  voltage.  This  has  much  application  in  light- 
ning-arrester design. 

It  should  be  noted  that  the  distribution  of  electrostatic 
stress  in  a  gap  can  be  simply  mapped  only  when  no  current 
is  passing.  For  instance,  the  field  between  filament  and 
plate  of  a  conducting  thermionic  tube  cannot  be  simply 
drawn.  The  presence  of  the  electrons  in  the  space  between, 
that  is,  the  space-charge  effect,*  causes  the  gradient  to  be  modi- 
fied and  concentrates  the  stress  near  the  positive  electrode. 

Similarly  in  a  glow  tube  which  is  ionized,  we  may  have  a 
space  charge.  When  a  molecule  is  ionized,  the  electron  and 
the  positive  ion  both  tend  to  move  out  of  the  field.  If 
the  gas  is  hydrogen,  however,  the  ion  weighs  about  1700 
times  as  much  as  the  electron.  The  charge  and  hence  the 
force  on  each,  are  the  same.  The  electrons  thus  move  more 
quickly  and  are  sooner  swept  out  of  the  ionized  space. 
The  preponderance  of  positive  ions  in  this  space  near  the 
anode  brings  a  concentration  of  gradient  nearer  to  the  cath- 
ode. In  a  glow  tube  the  greater  part  of  the  voltage  drop 
will  therefore  be  found  to  be  within  a  short  distance  of  the 
cathode.  This  effect  is  sometimes  modified  by  negative 
ions  which  are  formed  as  follows.  An  electron  in  its  flight 
often  attaches  itself  to  a  molecule  and  carries  it  along 
with  it.  This  constitutes  a  negative  ion,  which  is  as  heavy 
and  moves  as  slowly  as  a  positive  ion,  that  is,  as  a  mole- 
cule minus  an  electron. 

Air  on  the  surface  of  an  insulator  breaks  down  more  easily 

*  See  page  396  for  further  explanation  of  this  effect. 


CONDUCTION  THROUGH  GASES 


419 


than  air  in  a  space  between  separated  electrodes.  Greater 
creepage  distance  must  therefore  be  allowed  over  surfaces 
in  order  to  sustain  a  given  voltage.  This  is  probably  due 


FIG.  203o.  A  flash-over  on  a  wet  bushing  during  a  test.  A  potential 
of  305,000  volts  was  being  applied  when  this  picture  was  taken.  The 
General  Electric  Co. 

in  part  to  a  surface  layer  of  moisture  on  even  a  surface 
which  appears  dry.  The  appearance  of  such  a  flash-over  on 
a  wet  bushing  is  shown  in  Fig.  203a 

Prob.  9-12.  A  125-millimeter  diameter  sphere  is  situated  56 
millimeters  above  the  floor,  which  is  of  metal.  What  poten- 
tial will  be  necessary  between  sphere  and  floor  to  cause  a  spark 
to  pass?  Assume  standard  conditions. 


420      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  10-12.  What  potential  will  break  down  a  ball  gap  of 
spheres  of  25-centimeter  diameter,  spaced  9  centimeters,  when 
the  temperature  is  15°  C.  and  the  barometer  70  centimeters/ 

Prob.  11-12.  A  transmission  line  carries  power  at  110  kilo- 
volts  from  a  station  in  the  mountains  to  a  city  in  the  valley. 
If  the  same  insulators  are  used  throughout,  where  are  they  more 
likely  to  flash  over?  If  the  barometer  in  the  mountains  is 
60  centimeters  when  it  is  76  centimeters  in  the  valley,  how 
great  a  factor  is  thus  introduced?  If  it  is  also  colder  in  the 
mountains,  how  does  this  affect  flash-overs? 

114.  Corona.  It  does  not  always  follow  that  break  down 
of  air  at  one  point  between  electrodes  will  result  in  a  com- 
plete break  down.  When  the  layer  of  gas  adjacent  to  the 
electrode  is  ionized,  the  full  stress  is  brought  to  bear  on  the 
remaining  space.  Whether  the  stress  on  the  next  layer  of 
gas  of  this  remaining  space  is  thus  brought  to  the  critical 
value  depends  upon  the  shape  of  the  electrodes  and  their 
separation.  With  small  electrodes  far  apart,  it  will  often 
happen  that  in  the  new  distribution  of  stress  when  the  sur- 
face layer  of  gas  is  ionized,  there  will  be  no  new  point  further 
out  from  the  wire  where  the  gradient  is  sufficiently  raised 
to  continue  the  spread  of  the  discharge. 

In  such  cases  we  have  a  brush  discharge  or  a  corona. 
Such  a  case  occurs  particularly  on  the  wires  of  a  high-tension 
transmission  line.  For  the  voltage  at  which  many  of  these 
are  used,  the  gradient  at  the  surface  of  the  wire  is  above 
the  critical  gradient,  and  yet  the  configuration  is  such  that 
break  down  from  wire  to  wire  does  not  occur.  In  use  each 
wire  is  surrounded  by  a  luminous  envelope  of  ionized  air. 
The  production  of  ions  and  the  discharging  on  the  wire 
constitute  a  leakage  current  from  the  wire.  This  occasions 
a  loss  in  the  transmission  line.  Due,  however,  to  the  great 
advantages  of  the  use  of  high  voltages,  a  small  amount  of 
corona  loss  can  often  be  tolerated. 


Prob.  12-12.  The  leakage  current  from  each  wire  of  a 
transmission  line  due  to  corona  is  one  ampere,  the  line  being 
100  miles  long.  Assuming  two  wires  only  with  110,000  volts 


CONDUCTION  THROUGH  GASES  421 

applied,  how  much  loss  does  this  involve?  The  line  current 
at  full  load  is  100  amperes.  The  line  wires  have  a  resistance 
of  0.4  ohm  per  mile.  Dropping  the  voltage  to  88,000  volts 
would  stop  corona  entirely.  From  the  standpoint  of  efficiency 
of  transmission,  would  it  be  advisable  to  make  this  change? 

Prob.  13-12.  With  the  corona  of  Prob.  12-12,  how  many  ions 
per  second  are  being  discharged  on  each  centimeter  length  of 
line  wire,  assuming  each  ion  singly  charged? 

Prob.  14-12.  How  could  the  corona  in  Prob.  12-12  be  avoided 
without  decreasing  the  line  voltage? 

115.  Arcs.  It  requires  an  enormous  potential  gradient 
to  break  down  air  at  atmospheric  pressure,  but  once  the  air 
is  disrupted  and  a  spark  is  passed,  even  a  fairly  small  voltage 
can  maintain  a  current  across  the  gap. 

The  sudden  explosive  flow  of  current  when  air  is  broken 
down  by  a  high  potential  gradient  is  called  a  spark.  The 
steady  current  which  follows  at  low  potential  drop  is  called 
an  arc. 

The  arc  has  certain  distinctive  properties  which  cause  it 
to  differ  from  discharge  through  gas  at  low  pressure.  It 
is  confined  to  a  relatively  narrow  path  between  the  elec- 
trodes instead  of  being  uniformly  spread  out.  It  therefore 
produces  an  intense  heat  at  a  small  point  on  the  electrode. 
If  the  arc  is  long  continued,  this  point  is  brought  to  incandes- 
cence and  melted  or  even  vaporized.  When  this  occurs, 
the  molecules  of  vapor  are  charged  when  they  leave  the 
electrode  and  assist  in  carrying  the  current.  The  positive 
ions  when  they  bombard  the  incandescent  metal  cause  it 
to  emit  electrons  copiously.  There  is  also  a  thermionic 
emission  by  which  electrons  are  evaporated  from  the  cath- 
ode surface.  In  the  air  space  through  which  the  arc  passes, 
there  is  intense  heat  which  renders  the  air  molecules  easily 
ionized.  For  all  these  reasons,  an  arc  which  has  continued 
long  enough  to  heat  the  electrodes  is  easily  maintained  and 
large  amounts  of  current  can  pass  at  low  voltage.  Fifty 
volts  will  maintain  an  intense  arc  between  carbon  electrodes 
in  air. 


422      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

An  arc  has  a  negative  characteristic.  The  higher  the 
current  passed,  the  lower  is  the  drop  between  electrodes. 
The  reason  for  this  is  easily  seen.  The  heavier  the  current 
carried,  the  fatter  the  arc  becomes.  The  resistance  is  there- 
fore much  less  for  a  heavy  current.  In  fact,  the  conductivity 
of  the  arc  path  increases  at  a  greater  rate  than  simple  pro- 
portionality to  the  current. 

There  are  many  kinds  of  arcs.  Arcs  between  carbon 
electrodes  are  much  used  for  lighting.  In  such  a  case, 
most  of  the  light  comes  from  the  incandescent  electrodes 
themselves.  The  arc  stream  is  relatively  much  less  luminous. 
In  certain  arc  lights,  however,  known  as  flaming  arcs,  the 
light  from  the  arc  stream  itself  is  increased  by  impregnating 
the  carbon  electrodes  with  certain  salts. 

An  arc  light  must  be  operated  on  a  constant-current  cir- 
cuit or  with  a  resistance  in  series  for  stabilizing.  Due  to 
its  negative  characteristic,  an  arc  connected  to  a  constant- 
potential  circuit  will  be  very  unstable.  The  more  current 
it  takes,  the  less  is  its  drop,  and  it  hence  acts  like  a  complete 
short  circuit.  A  resistance  in  series  will  give  a  rising  charac- 
teristic for  lamp  and  resistance  and  hence  stabilize  it.  For 
a  street-lighting  circuit,  arcs  are  connected  in  series  and  sup- 
plied by  a  transformer  which  automatically  maintains  a 
constant  current,  thus  avoiding  possible  instability. 

The  mercury-arc  lamp  is  also  used  for  lighting.  In  this 
case  the  arc  passes  between  a  solid  metal  electrode  and  a 
pool  of  liquid  mercury,  or  between  two  mercury  pools. 
The  gas  present  is  simply  the  mercury  vapor.  To  start 
the  lamp,  it  is  tilted  until  a  stream  of  mercury  passes  over 
and  touches  the  opposite  electrode.  When  this  stream  breaks, 
the  arc  follows.  Since  the  mercury  vapor  is  at  low  pressure, 
the  arc  diffuses  throughout  the  tube.  It  concentrates,  how- 
ever, at  a  small  spot  on  the  pool  of  mercury.  The  intense 
heat  at  this  point,  and  the  charged  vapor  and  electrons 
thus  produced,  allow  the  passage  of  current  at  a  low  vol- 
tage. 


CONDUCTION  THROUGH  GASES 


423 


In  the  mercury  arc,  there  is  an  incandescent  spot  on  the 
pool  of  mercury,  but  none  on  the  other  electrode.  The 
electrons  evaporated  from  this  spot  are  the  principal  means 
by  which  the  arc  is  sustained.  Current  can  therefore  pass  in 
only  one  direction  through  a  mercury  arc  between  the  mer- 
cury and  a  solid  electrode.  It  may  thus  be  used  as  a  recti- 


FIG.  204.     Welding  the  ends  of  hot-rolled  steel  rings.   Westinghouse 
Electric  and  Mfg.  Co. 

fier.  A  mercury-arc  rectifier,  for  changing  alternating  into 
direct  current,  is  quite  similar  in  operation  to  the  mercury 
arc  used  for  lighting.  It  is  designed,  however,  with  a_differ- 
ent  purpose  in  view. 

Arcs  are  also  much  used  for  welding  metals.     In  the  pro- 
cess which  is  most  used  for  welding  iron  and  steel  as  shown 


424      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

in  Fig.  204,  one  electrode  is  the  material  to  be  welded  and 
the  other  is  an  iron  rod  held  in  the  hand.  The  rod  is  touched 
to  the  work  to  start  the  arc,  and  then  withdrawn  slightly 
to  form  an  arc  about  one-fourth  of  an  inch  long.  Only  30  to 
50  volts  are  required  to  maintain  the  arc.  A  current  of  50 
to  150  amperes  is  used,  depending  upon  the  nature  of  the 
work.  The  arc  raises  the  material  to  be  welded  to  fusion 
temperature.  Small  globules  of  the  ' '  pencil ' '  or  electrode  held 
in  the  hand  are  dropped  in  molten  condition  into  the  weld. 
Considerable  skill  is  required  in  order  to  produce  uniform 
work. 

Besides  the  useful  applications  of  arcs,  there  are  many 
places  where  they  must  be  dealt  with  by  the  electrical  en- 
gineer in  connection  with  switching  and  line  protection. 
Their  nature  should  therefore  be  thoroughly  understood. 

It  has  been  mentioned  that  an  arc  has  a  negative  charac- 
teristic. In  Fig.  205,  page  430,  is  shown  the  characteristic  for 
an  arc  in  air  between  metal  electrodes.  It  will  be  noted  that 
as  the  current  is  decreased,  a  point  is  reached  where  the 
voltage  rises  very  rapidly  indeed.  It  requires  a  certain 
current  to  maintain  an  arc.  If  the  current  falls  below  this 
value,  the  arc  will  suddenly  go  out.  The  current  necessary 
to  maintain  the  arc  depends  upon  its  length,  the  electrode 
material,  the  electrode  temperature  and  several  other  factors. 

When  a  switch  is  opened  in  a  power  circuit,  an  arc  follows 
the  separation  of  the  blades.  If  the  potential  of  the  source 
is  high  or  if  there  is  much  inductance  in  the  circuit,  the  arc 
may  be  drawn  out  very  long,  in  fact,  several  feet  on  high- 
voltage  power  circuits.  If  the  source  is  capable  of  main- 
taining the  arc  at  its  maximum  length,  it  will  continue  until 
interrupted  in  some  manner,  or  until  it  has  fused  and  de- 
stroyed the  switch.  If  the  arc  is  drawn  out,  however,  until 
the  current  passing  becomes  too  small  to  maintain  it,  the 
arc  will  suddenly  go  out.  It  is  this  sudden  extinguishing 
of  the  arc  which  produces  a  rapid  change  of  current  and  a 
high  potential  when  an  inductive  circuit  is  opened.  Thus 


CONDUCTION  THROUGH  GASES        425 

when  the  field  circuit  of  a  large  machine  is  suddenly  opened 
by  a  switch,  it  is  not  the  instant  when  the  contacts  separate 
that  is  dangerous,  for  a  hundred  volts  or  so  will  start  and 
maintain  the  arc,  but  it  is  the  instant  when  the  arc  breaks. 

An  arc  stream  which  is  situated  in  a  magnetic  field  is 
forced  sideways  in  exactly  the  same  way  that  a  wire  carry- 
ing current  would  be  acted  upon.  The  cause  of  the  action 
is  the  same  in  both  cases,  namely  the  tendency  of  a  current- 
carrying  conductor  in  a  magnetic  field  to  be  deflected  side- 
ways. The  direction  in  which  the  arc  will  move  can  be 
determined  by  the  left-hand  rule,  just  as  in  the  case  of  a 
motor. 

This  principle  is  made  use  of  in  magnetic  blowouts  as 
applied  to  trolley-car  controllers,  contactors,  etc.  When 
the  contact  points  separate,  there  is  a  magnetic  field  in  the 
space  between  them  which  forces  the  arc  out.  This  field 
is  usually  produced  by  a  coil  connected  in  series  with  the 
circuit  to  be  opened.  Then  the  heavier  the  current  to  be 
opened,  the  stronger  will  be  the  field  acting  to  blow  the  arc. 
The  movement  of  the  arc  in  the  field  accomplishes  two  things. 
First,  the  arc  is  lengthened  and  moved  out  to  where  it  will  do 
no  harm.  A  longer  arc,  we  have  seen,  will  break  more  eas- 
ily. Second,  it  is  moved  over  to  a  cool  surface.  It  has 
been  shown  that  one  condition  necessary  for  the  mainte- 
nance of  an  arc  at  low  voltage  is  an  incandescent  spot  on 
the  surface  of  the  electrode.  If  the  arc  is  moved  off  this 
spot,  it  will  have  a  powerful  tendency  to  go  out.  This 
second  point  is  not  always  fully  appreciated. 

Prob.  15-12.  Using  the  curve  of  Fig.  205,  draw  the  char- 
acteristic for  the  combination  of  this  arc  in  series  with  10  ohms 
resistance.  Will  the  combination  be  stable  at  6  amperes? 
What  is  the  apparent  resistance  of  the  combination  at  this 
current  value?  At  8  amperes? 

Prob.  16-12.  Make  a  sketch  showing  a  switch  with  mag- 
netic blowout,  indicating  the  coil  connections  and  the  direction 
in  which  the  arc  will  be  blown.  Will  the  action  be  correct  if 
the  current  to  be  broken  is  reversed  in  direction? 


SUMMARY    OF   CHAPTER   XII 

THE  FLOW  OF  ELECTRICITY  THROUGH  GASES  is  not 
governed  simply  by  Ohm's  law. 

BY  THERMIONIC  CONDUCTION  is  meant  the  conduction 
of  electricity  by  electrons  which  are  evaporated  from  a  hot 
conductor.  The  hot  conductor,  together  with  a  cold  plate,  is 
placed  in  a  vacuum  bulb  and  a  voltage  applied  between  the 
hot  conductor  and  the  cold  plate. 

RICHARDSON'S  LAW.  The  amount  of  current  which 
passes  in  a  thermionic  tube  with  the  hot  filament  at  a  given  tem- 
perature increases  as  the  voltage  between  filament  and  plate  in- 
creases up  to  a  certain  point,  where  saturation  occurs.  The 
saturation  point  depends  upon  the  temperature  of  the  hot  fila- 
ment. The  saturation  current  is  given  by  the  equation 

i        K 

is  =  a  T2  e~  f    . 

AN  AMPLIFIER  OR  REPEATER  is  made  by  placing  a 
grid  between  the  hot  filament  and  cold  plate.  The  amount  of 
current  passed  between  the  hot  filament  and  cold  plate  is  caused 
to  vary  through  wide  ranges  by  means  of  slight  changes  in  the 
potential  on  the  grid.  Thus  the  small  telephone  currents  may 
be  made  to  cause  the  grid  potential  to  vary  and  in  this  way  con- 
trol the  flow  of  much  larger  currents  in  the  circuit  of  the  plate. 
The  output  current  will  thus  vary  in  the  same  manner  as  the 
input  current.  This  is  the  action  of  the  telephone  repeater. 

X-RAYS  ARE  SHORT  WAVES  set  up  at  the  target  of  a  bulb 
when  electrons  at  very  high  velocity  impinge  upon  the  metal 
surface  of  the  target.  These  X-rays  can  penetrate  substances 
impervious  to  ordinary  light  waves. 

THE  VELOCITY  OF  ELECTRONS  can  be  found  from  the 
following  equation 


v  = 

This  equation  does  not  [hold  for  velocities  approaching  that 
of  light  unless  we  consider  the  value  of  m  to  increase  with  the 
velocity. 

426 


CONDUCTION   THROUGH  GASES  427 

CONDUCTION  MAY  TAKE  PLACE  IN  A  TUBE  CONTAIN- 
ING GAS  at  low  pressure.  Free  electrons  in  the  gas  are  set 
in  motion  by  the  electric  force.  These  electrons  hit  molecules 
and  ionize  them  by  knocking  electrons  away  from  them.  During 
ionization  a  molecule  emits  a  spurt  of  light  the  color  of  which 
depends  upon  the  nature  of  the  molecules.  The  ions  upon 
hitting  the  cathode  knock  other  electrons  from  it,  keeping  up  the 
supply  of  free  electrons. 

TO  PRODUCE  A  CUMULATIVE  DISCHARGE  the  electric 
potential  must  be  sufficiently  high  to  give  the  electrons  enough 
acceleration  and  the  MEAN  PATH  of  the  electron  must  be 
long  enough  to  allow  the  electron  to  acquire  sufficient  velocity 
to  break  up  a  molecule  with  which  it  may  collide  and  thus 
ionize  it. 

SPARKS  ARE  PRODUCED  when  the  gas  pressure  is  so 
great  as  to  require  a  comparatively  high  potential  to  give  ion- 
izing velocity  to  the  electrons.  When  the  potential  is  reached 
at  one  spot  of  the  gap,  the  air  throughout  the  gap  becomes  sud- 
denly ionized  and  a  spark  passes. 

THE  SPARKING  DISTANCE  in  air  depends  upon  the  atmos- 
pheric pressure  and  the  temperature  as  well  as  upon  the  shape 
of  the  terminals,  pointed  terminals  requiring  less  potential. 

CORONA  IS  PRODUCED  when  the  ionization  of  one  sec- 
tion of  gas  between  electrodes  does  not  result  in  raising  the 
potential  gradient  of  any  other  section  to  the  ionizing  point. 
This  is  often  the  condition  on  high-voltage  transmission  lines. 

ARCS  ARE  CHARACTERIZED  BY  THE  VAPORIZATION 
of  electrodes  and  ionization  of  the  vapor.  Arcs  differ  from 
sparks  in  being  confined  to  the  comparatively  narrow  path  of 
vapor  flow  between  electrodes  which  may  be  solid,  as  in  carbon- 
arc  lamps,  or  liquid  as  in  the  mercury-arc  lamp. 


PROBLEMS   ON   CHAPTER   XII 

Prob.  17-12.  In  the  Western  Electric  standard  repeater, 
the  filament  has  a  coating  of  the  oxides  of  barium  and  strontium. 
The  area  of  the  coating  is  approximately  95  square  millimeters. 
The  value  of  a  in  Richardson's  equation  is  2.0  X  105.  At  a 
temperature  of  750°  C.  the  saturation  current  is  0.04  ampere. 
What  is  the  value  of  K  in  Richardson's  equation  for  this  sub- 
stance? 

Prob.  18-12.  If  the  filament  of  the  repeater  in  Prob.  17-12  is 
heated  to  850°  C.,  what  will  the  saturation  current  become? 
One  of  the  good  features  of  the  oxides  composing  the  coating  of 
the  filament  in  these  repeaters  is  the  high  current  of  saturation 
at  comparatively  low  temperatures. 

Prob.  19-12.  Determine  the  path  followed  by  a  charged 
particle,  moving  with  a  velocity  u  in  a  magnetic  field  of  strength 
B.  The  charge  on  the  particle  is  q  and  the  mass  is  m.  The 
angle  between  the  initial  direction  and  that  of  the  magnetic  flux 
is  90°.  Neglect  the  effect  of  gravity. 

Prob.  20-12.  The  copper  target  of  an  X-ray  tube  weighs 
200  grams.  When  the  tube  is  first  started  the  temperature 
of  the  target  rises  at  the  rate  of  8°  C.  per  second.  If  10  milli- 
amperes  are  being  passed  through  the  tube,  how  fast  are  the 
electrons  moving  when  they  arrive  at  the  target?  Assume  the 
mass  of  the  electron  to  be  constant. 

Prob.  21-12.  A  thermionic  valve  has  been  tested  for  its 
saturation  current  at  two  filament  currents.  The  results  ob- 
tained are  as  follows: 

Filament  current  Saturation  current 

0.90  0.0105 

1.05  0.0455 

The  value  of  a  in  Richardson's  equation  for  this  tube  is  6.0  X  104. 
Assuming  that  the  heat  radiated  from  the  filament  varies  as  the 
fourth  power  of  the  absolute  temperature,  and  that  the  re- 
sistance of  the  filament  is  not  appreciably  changed  by  the  change 
in  temperature,  determine  the  value  of  k  in  the  saturation- 
current  formula,  and  also  determine  the  saturation  current 
at  a  filament  current  of  1.1  amperes. 

428 


CONDUCTION  THROUGH  'GASES  429 

Prob.  22-12.  Determine  the  absolute  temperature  of  the 
filament  in  each  of  the  cases  above.  Plot  a  curve  showing  the 
relation  between  filament  current  and  saturation  current  for 
the  region  including  the  above  two  points.  Plot  another  curve 
showing  the  relation  between  saturation  current  and  absolute 
temperature  for  the  same  region.  If  necessary,  compute  a 
few  other  points,  in  order  to  get  a  good  curve. 

Prob.  23-12.  If  the  temperature  coefficient  of  resistance  of 
a  filament  is  taken  as  c,  and  the  heat  radiated  from  it  is  taken 
as  k  times  the  fourth  power  of  the  absolute  temperature,  de- 
rive equations  for  the  following: 

(a)  relation  between  the  filament  current  and  the  absolute 
temperature ; 

(6)  relation  between  power  input  and  resistance  of  filament. 

(c)  Plot  a  curve  for  each  of  the  above  cases  on  the  assumption 
that  c  =  0.009  and  k  =  5  X  lO"11. 

Prob.  24-12.  The  saturation  current  of  a  tube  maintains  the 
plate  at  a  constant  temperature  of  400°  C.,  with  a  fall  in  potential 
between  filament  and  plate  of  180  volts.  To  what  must  the 
plate  potential  be  increased  if  the  temperature  is  to  be  raised 
to  650°  C.,  the  filament  current  remaining  constant? 

Prob.  25-12.  The  plate  in  the  tube  in  Prob.  21-12  is  made  of 
the  same  material  as  the  filament,  and  its  area  (on  the  filament 
side)  is  100  times  the  filament  area.  The  plate  current  bom- 
bardment maintains  it  at  a  temperature  of  350°  C.  If  the  direc- 
tion of  the  plate  potential  is  now  reversed,  find  the  current  which 
will  flow  from  plate  to  filament.  What  temperature  must  the 
plate  assume  if  this  current  is  to  be  one-hundredth  of  the  sat- 
uration current  from  plate  to  filament,  when  the  filament 
current  is  1.0  ampere? 

Prob.  26-12.  Using  the  mean  free  path  of  an  electron  in  air 
at  normal  atmospheric  pressure  found  in  Prob.  7-12,  determine 
the  potential  which  must  exist  on  the  surface  of  a  sphere  to 
start  ionization  at  its  surface.  Assume  an  ionization  velocity 
of  2  X  108  centimeters  per  second. 

Prob.  27-12.  A  transmission  line  is  50  miles  long  and  has 
a  resistance  of  0.403  ohm  per  thousand  feet.  The  maximum 
allowable  carrying  capacity  of  this  wire  is  50  amperes.  At 
90.000  volts  at  the  receiving  end, the  corona  loss  on  this  line  is 
5  kw.  per  mile,  and  at  110,000  volts  the  loss  is  35  kw.  per  mile. 
The  corona  loss  can  be  assumed  to  vary  according  to  the  law 

p  =  c(V  -  F0)2, 


430      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

where  V  is  the  line  voltage,  and  c  is  a  constant,  dependent  on 
spacing,  diameter  of  wire  and  atmospheric  conditions.  F0  is  the 
voltage  at  which  the  corona  first  appears.  If  power  costs  2j 
cents  per  kilowatt-hour  to  generate  and  sells  for  4  cents  per  kilo- 
watt-hour, compute  the  theoretical  voltage  at  which  this  line 
will  operate  when  bringing  the  greatest  income.  Neglect  the 
effect  of  line  drop  upon  corona  loss.  (In  practice  the  operating 
voltage  would  be  determined  by  many  other  additional  con- 
siderations.) 


-200 


-150 


-100 


Amperes 

FIG.  205.    Relation  between  current  and  voltage  of  an  arc 
between  metal  electrodes. 

Prob.  28-12.  An  arc  between  metal  electrodes  has  the  charac- 
teristic shown  in  Fig.  205. 

(a)  Plot  a  curve  using  as  abscissas  the  values  of  current,  and 
as  ordinates,  the  values  of  the  minimum  resistance  that  must  be 
placed  in  series  with  the  arc  to  give  stable  operation. 

(6)  Plot  a  curve  of  voltage  across  both  arc  and  resistance  for 
the  conditions  in  (a),  using  the  same  abscissas. 


CHAPTER  XIII 
DIELECTRICS 

An  electric  current  flows  whenever  electrons  are  caused 
to  move.  When  a  voltage  is  applied  to  the  ends  of  a  wire, 
the  free  electrons  in  the  conductor  wander  along  the  wire  and 
there  is  metallic  conduction.  If  the  wire  is  broken  and 
the  two  ends  placed  in  an  electrolyte,  a  current  continues 
to  flow.  Electrons  from  the  wire  become  attached  to  ions 
in  the  solution  and  move  with  them  through  the  electrolyte. 
Even  if  there  is  an  air  gap  in  the  wire,  current  can  still  flow 
through  a  spark  or  an  arc  if  the  potential  is  high  enough,  con- 
duction in  this  case  being  due  to  electrons,  gaseous  ions  and 
charged  particles  of  metal  which  move  from  one  electrode 
to  the  other.  Thus  in  order  to  have  a  current,  we  must 
in  all  cases  have  electric  charges  free  to  move. 

All  substances  are  constructed  of  atoms,  and  all  atoms 
consist  of  electrons  attached  in  some  mysterious  manner  to 
a  charged  nucleus.  If  the  substance  is  a  metal,  the  elec- 
trons are  continually  interchanging  between  atom  and 
atom,  and  even  a  small  force  impressed  upon  them  will 
cause  a  movement  along  the  wire.  If  the  substance  is  an 
insulator  such  as  glass,  however,  there  are  about  as  many 
electrons  present,  but  they  are  securely  attached  to  the 
atoms  and  cannot  get  away  under  the  action  of  moderate 
electric  forces.  The  application  of  an  electromotive  force 
of  small  value  to  an  insulator  hence  produces  practically 
no  steady  current. 

116.  Dielectric  Strength.  A  substance  which  will  con- 
duct practically  no  steady  current  of  electricity  at  all  under 
a  moderate  applied  electromotive  force  is  called  an  insulator 
or  a  dielectric.  The  principal  insulators  used  in  electrical 
work  are:  first,  mica  and  the  vitreous  materials  such  as  glass, 

431 


432      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

porcelain  and  so  on;  second,  rubber  and  its  derivatives; 
third,  paper  and  moulded  composition;  and  fourth,  the  oils 
and  varnishes. 

In  a  dielectric,  we  have  said  that  the  electrons  are  se- 
curely attached  to  the  atoms  and  hence  cannot  flow  along 
the  material  when  a  moderate  electromotive  force  is  applied. 
However,  if  the  electromotive  force  is  raised  to  a  sufficient 
value,  the  electrons  can  possibly  be  torn  loose  and  the  ma- 
terial thus  broken  down  or  disrupted  in  exactly  the  same 
way  that  air  is  broken  down.  When  an  insulator  is  thus 
punctured,  the  electrons  are  torn  forcibly  loose  from  the 
atoms.  The  potential  gradient  necessary  to  accomplish 
this  is  very  high  and  depends  upon  the  material.  The 
potential  gradient  necessary  to  disrupt  an  insulation  is 
called  its  dielectric  strength. 

As  in  the  case  of  air,  it  is  not  necessary  that  a  high  poten- 
tial gradient  be  produced  at  all  points  in  the  body  of  the 
material  in  order  to  break  down  the  insulator.  It  is  suffi- 
cient that  at  one  point  the  potential  gradient  shall  exceed 
the  dielectric  strength.  In  such  a  case,  this  point  will 
break  down,  thus  bringing  the  whole  potential  to  bear 
on  the  remaining  insulation  and  so  breaking  it  down  in  turn. 
Thus  a  given  slab  of  insulating  material  may  stand  a  certain 
voltage  satisfactorily  when  it  is  applied  between  flat  elec- 
trodes held  against  each  surface,  whereas  it  will  soon  break 
down  if  the  same  potential  is  applied  between  points  on 
each  side  of  the  slab. 

Glass  is  one  of  the  best  dielectrics  from  an  electrical 
standpoint.  It  has  a  dielectric  strength  of  about  100,000 
volts  per  centimeter.  Mechanically,  it  is,  of  course,  open  to 
considerable  objection.  Porcelain  has  nearly  the  same 
dielectric  strength  and  is  much  stronger  mechanically. 
For  places  where  an  insulator  of  small  thickness  is  desired, 
mica  is  much  used.  Its  dielectric  strength  is  about  one- 
half  that  of  glass. 

The  dielectric  strength  of  a  material  varies  with  a  great 


DIELECTRICS  433 

many  factors.  For  instance,  any  such  material  as  oil  tends 
to  absorb  moisture,  and  the  percentage  of  moisture  content 
greatly  affects  the  dielectric  strength.  This  is  particularly 
true  of  transformer  oil.  Even  one-tenth  of  one  percent  of 
moisture  dissolved  in  the  oil  will  reduce  the  dielectric  strength 
to  less  than  ten  percent  of  full  value.  The  temperature  also 
has  a  large  effect.  In  general,  the  higher  the  temperature 
of  the  material,  the  less  will  be  its  dielectric  strength.  Fi- 
nally, it  should  be  mentioned  that  the  length  of  time  during 
which  the  voltage  is  applied  to  the  specimen  is  a  great  fac- 
tor, —  certain  substances  that  will  stand  up  for  one  minute 
under  a  voltage  of  125,000  volts  per  centimeter  will  break 
down  under  80,000  volts  per  centimeter  applied  for  a  half  hour. 

In  practice,  in  constructing  insulators  great  care  must  be 
used  that  no  air  pockets  are  left  in  the  material.  If  there  is 
a  bubble  of  air  in  a  dielectric  which  is  subjected  to  high 
stress,  the  air  in  the  bubble  may  be  broken  down  although 
the  material  itself  is  far  below  puncture  voltage.  In  such 
a  case,  ionization  produced  in  the  bubble  results  in  a  great 
deal  of  heating.  This  heats  the  insulator  locally  and  often 
causes  it  to  break  down  where  it  would  otherwise  have  been 
fully  strong  enough  for  the  conditions  at  hand. 

The  proper  design  of  an  insulator  brings  in  many  factors. 
For  instance,  in  designing  an  insulator  for  a  transmission 
line,  we  must  first  construct  it  so  that  there  is  sufficient 
dielectric  strength  of  the  material  to  avoid  puncture  under 
the  maximum  voltage  gradient  which  will  occur.  Second, 
it  must  have  sufficient  surface  so  that  it  will  not  break 
down  between  wire  and  ground  over  the  surface  of  the  ma- 
terial. This  is  the  reason  that  insulators  are  constructed 
with  a  deeply  corrugated  surface  or  with  petticoats,  as 
shown  in  Fig.  206.  Also  these  petticoats  must  be  of  such 
shape  that  when  it  rains,  there  will  be  enough  surface  of 
the  insulator  left  dry  to  stand  the  applied  potential.  In 
addition,  of  course,  the  insulator  must  be  strong  enough  to 
stand  the  mechanical  stresses  which  are  brought  to  bear 


434       PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


upon  it,  and  these  are  often  high  when  a  heavy  transmission 
wire  is  being  supported.  Finally,  there  must  be  no  point 
close  to  the  insulator  at  which  the  air  is  highly  ionized,  and 

no  bubbles  rof  air 
in  the  material  of 
which  it  is  con- 
structed. 

Prob.  1-13.  Bake- 
lite  prepared  for 
•the  electrical  trade 
may  have  a  dielec- 
tric strength  of 
15,000  volts  per 
millimeter.  In  a 
certain  machine  it 
was  desired  to  re- 

FIG.  206.     An  insulator  having  petticoats  to    place  some  mica  in- 

'( increase  the  length  of  dry  leakage  path.    Wes-    sulating  sheetslmil- 
tinghouse  Electric  &  Manufacturing  Co.  limeter  thick  with 

bakelite     sheets. 

How  thick  should  the  bakelite  sheets  be?     Assume  that  the 

same  voltage  will  be  applied  in  each  case  between  the  surfaces 

of  the  sheets. 

Prob.   2-13.     For  rubber   insulation  the  relative   dielectric 

strengths  for  various  times  of  electrification  are  given  in  the 

following  table. 


Time  of  electrification 
in  minutes 

1 

3 

5 

10 

15 
30 


Relative  dielectric 
strength 

180 
110 
100 

90 

85 

80 


The  5-minute  factory  test  of  No.  00  rubber-insulated  cables 
having  -fa  inch  insulation  is  11.5  kilovolts.  What  voltage  will 
these  cables  safely  stand  for  1  minute? 

117.   Condenser  Action.     We  now  come  to  a  very  im- 
portant matter,  —  the  subject  of  electric  condensers. 


DIELECTRICS 


435 


i 

a 

1 

s             -4 

-B                 <, 

u 

We  have  seen  that  the  electrons  in  a  dielectric  are  fixed  in 
position  by  being  attached  to  specific  atoms.  However, 
the  bonds  which  are 
holding  the  electrons  to 
the  atoms  are  more  or 
less  flexible.  It  is  true 
that  with  only  mod- 
erate electromotive 
forces  it  is  impossible 
to  break  these  bonds 
and  cause  the  electrons 
to  move  through  the 
material.  However, 
any  electromotive  force 
whatever  will  somewhat 
stretch  the  bonds  and 
cause  the  electrons  to 
move  slightly  in  posi- 
tion. 

Thus  in  Fig.  207,  suppose  that  D  is  a  slab  of  dielectric 
material,  and  that  A  and  B  are  metal  electrodes  placed 
against  its  surface.  The  material  D  is  made  up  of  an 
enormous  number  of  atoms  with  attached  electrons.  Let  us 
represent  one  of  these  electrons  by  q,  and  represent  the  bond 
by  which  it  is  attached  to  the  atom  by  the  short  straight 
line.  Suppose  that  a  potential  is  applied  between  A  and 
B,  and  that  A  is  made  negative.  The  electron  will  now  be 
repelled  by  A  and  attracted  by  B.  It  will  therefore 
move  into  some  such  position  as  is  shown  by  the  dotted  line. 
While  it  is  thus  moving,  we  have  an  electric  current  through 
the  dielectric.  Note,  however,  that  this  current  continues 
for  only  an  instant.  If  we  suddenly  apply  the  voltage  be- 
tween A  and  B,  there  will  be  a  small  instantaneous  current 
as  the  electrons  move  slightly  in  their  positions  in  the  atom, 
but  this  current  will  soon  die  out,  and  then  after  a  time  there 
will  be  zero  current  as  long  as  the  applied  voltage  is  steady. 


FIG.  207.  A  simple  condenser,  consist- 
ing of  a  dielectric  D  held  between  two 
electrodes  A  and  B. 


436      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Moreover,  if  the  voltage  between  A  and  B  is  removed, 
the  electrons  by  reason  of  the  elasticity  of  their  bonds  will 
return  to  their  previous  positions.  That  is,  we  get  a  cur- 
rent on  applying  a  voltage,  and  a  current  in  the  opposite 
direction  upon  removing  the  voltage.  In  fact,  the  effects 
produced  in  these  two  cases  are  exactly  equal  and  opposite. 

In  general,  whenever  the  voltage  impressed  upon  a  die- 
lectric is  changed,  there  will  be  a  current  in  the  dielectric 
which  we  call  a  displacement  current,  and  which  is  due  to 
the  slight  movement  of  the  electrons  in  their  positions  in 
the  atoms.  When  the  applied  voltage  is  steady,  there  is  no 
current.  When  the  voltage  is  varying,  a  current  is  pro- 
duced which  is  proportional  to  the  rate  of  change  of  the 
voltage. 

It  has  been  checked  experimentally  that  the  current  pro- 
duced in  a  dielectric  is  proportional  to  the  rate  of  change  of 
voltage.  Thus,  if  in  Fig.  207  a  voltage  is  gradually  applied 
between  A  and  B,  and  in  such  a  way  that  it  increases  from 
zero  to  100  volts  in  one  second,  a  certain  current  will  be  pro- 
duced which  we  will  call  7i.  If  now  the  rate  of  change  of 
voltage  is  increased,  the  current  will  be  increased.  The 
current  depends  simply  upon  the  rate  of  change  of  voltage. 
For  example,  if  the  voltage  between  A  and  B  is  increased  to 
100  volts  in  one-half  a  second,  the  current  produced  in  this 
interval  will  be  on  the  average  27 1,  and  so  on.  This  pro- 
portionality may  be  expressed  by  means  of  the  equation 

~de  f^ 

*  =  CdT  (1> 

The  proportionality  factor  C  is  called  the  capacitance, 
and  an  arrangement  consisting  of  a  dielectric  between  two 
electrodes  is  called  a  condenser. 

In  general,  then,  the  current  produced  in  a  condenser  is 
equal  to  the  product  of  its  capacitance  times  the  rate  of 
change  of  the  voltage.  This  may  be  compared  with  an 
inductance  coil,  where  we  saw  that  the  voltage  produced  was 


DIELECTRICS  437 

equal  to  the  inductance  of  the  coil  times  the  rate  of  change 
of  the  current. 

The  unit  of  capacitance  in  practical  units  is  the  farad. 
A  condenser  has  a  capacitance  of  one  farad  when  a  current 
of  one  ampere  will  flow  through  it  upon  the  application  of  a 
voltage  changing  at  the  rate  of  one  volt  per  second.  It 
has  been  found,  however,  that  this  unit  of  capacitance 
is  usually  much  too  large  for  engineering  use.  In  practical 
computations  it  is  therefore  much  more  convenient  to  use 
one-millionth  of  this  unit,  called  a  microfarad.  A  micro- 
farad is  the  capacitance  of  a  condenser  in  which  a  micro- 
ampere will  be  produced  by  a  voltage  changing  at  the  rate  of 
one  volt  per  second. 

We  may  also  write  equation  (1)  in  c.g.s.  units.  The 
c.g.s.  unit  of  capacitance  in  the  electromagnetic  system 
is  called  the  abfarad.  The  capacitance  of  a  condenser  in 
abfarads  is  equal  to  the  abamperes  of  current  produced 
divided  by  the  rate  of 
change  of  potential  in  ab- 
volts  per  second. 

Prob.  3-13.  A  condenser 
with  a  capacitance  of  1  micro- 
farad is  connected  as  shown 
in  Fig.  208.  If  the  slide  wire 
is  120  inches  long  and  of  a 
total  resistance  of  5  ohms  O-1  ohm 

and  if  the  slide  is  being  moved     FIG.  208.    As  the  slider  moves  to 
to  the  right  at  the  rate  of  5         the  right,  the  voltage  across  the 
feet  per  second,  what  current         condenser  drops, 
will  flow  in  the  condenser? 

Prob.  4-13.  How  much  would  you  modify  the  result  in 
Prob.  3-13  if  the  slide  wire  were  replaced  by  another  with  a 
resistance  of  one  ohm? 

Prob.  5-13.  What  is  the  capacitance  of  a  condenser  in  which 
a  current  of  0.012  ampere  flows  when  the  voltage  across  its  ter- 
minals rises  at  a  uniform  rate  from  100  to  110  volts  in  0.08  second? 

Prob.  6-13.  What  maximum  current  will  flow  in  a  condenser 
of  2.2  microfarads  when  an  alternating  pressure  of  150  volts 


r 


1.5  volts 


438      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

(maximum  value)  is  applied  to  its  terminals?     The  frequency 
of  the  voltage  is  60  cycles  per  second. 

118.  Dielectric  Constant.  The  capacitance  of  such  an 
arrangement  as  is  shown  in  Fig.  207  has  been  found  to  de- 
pend upon  the  material  of  the  dielectric  used.  If  B  is  a 
sheet  of  glass,  the  capacitance  will  be  found  to  be  about 
twice  what  it  is  when  B  is  composed  of  mica. 

The  dielectric  constant  of  a  material  is  a  measure  of  its 
effectiveness  when  used  in  a  condenser,  just  as  the  per- 
meability of  a  material  is  a  measure  of  the  effectiveness 
of  the  material  for  use  in  a  magnetic  circuit.  It  will  be  re- 
membered from  the  study  of  magnetic  circuits  that  the  per- 
meability of  iron  is  a  number  of  thousands,  and  of  other 
materials  much  lower,  but  that  the  permeability  of  even  air 
is  unity.  In  the  same  way,  when  we  study  condensers  we 
find  that  while  the  dielectric  constant  of  most  insulating 
materials  is  high  compared  with  that  of  air,  yet  air  or  even 
a  vacuum  has  a  dielectric  constant;  that  is,  both  will 
give  condenser  effects  when  between  parallel  plates.  When 
the  dielectric  of  a  condenser  is  stressed,  we  can  visualize 
the  movement  of  electrons  which  constitutes  the  displace- 
ment current.  It  has  never  been  satisfactorily  explained, 
however,  why  there  is  still  a  current  when  the  dielectric  is 
removed  and  replaced  by  a  vacuum.  Such,  however,  has 
been  found  to  be  the  case  experimentally.  There  is  by  no 
means  as  much  difference  in  the  dielectric  constant  of  vari- 
ous materials,  however,  as  there  is  in  the  permeability. 

The  dielectric  constant  of  air  is  so  nearly  the  same  as  that 
of  a  vacuum  that  it  is  not  usually  necessary  to  distinguish 
between  them.  We  may  therefore  define  the  dielectric 
constant  of  a  dielectric  material  as  the  capacitance  of  a 
condenser  constructed  with  this  material  as  dielectric  di- 
vided by  the  capacitance  of  exactly  the  same  condenser  when 
air  replaces  the  dielectric  material. 

Glass  has  a  dielectric  constant  ranging  between  five  and 
ten,  depending  upon  the  grade  of  the  glass.  Mica  has  about 


DIELECTRICS  439 

half  of  this.  Some  liquids  have  high  dielectric  constants, 
that  of  glycerine  being  about  fifty-six  and  of  picric  acid  about 
eighty.  Very  pure  water  is  an  insulator  and  has  a  die- 
lectric constant  which  can  be  measured.  It  is  in  the  neigh- 
borhood of  eighty.  Table  VI  in  the  appendix  gives  the 
dielectric  constants  for  materials  commonly  used  in  con- 
densers. 

Prob.  7-13.  A  certain  condenser  using  glycerine  as  the 
dielectric  has  a  capacitance  of  1.84  microfarads.  The  pressure 
across  it  is  varies  from  8  volts  to  24  volts  in  0.003  second. 
What  average  current  will  flow  in  this  condenser  under  this 
condition,  if  the  glycerine  is  replaced  by  acetone? 

119.  Parallel-Plate  Condensers.  A  parallel-plate  con- 
denser may  be  constructed  as  shown  in  Fig.  207  or,  in  order 
to  increase  the  amount  of  capacitance,  as  shown  in  Fig.  209. 


FIG.  209.    A  parallel-plate  condenser. 

Such  condensers  may  be  built  up  of  sheets  of  metal  and  plates 
of  glass.  They  may  be  constructed  of  paraffined  paper  and 
tinfoil.  The  construction  depends,  of  course,  upon  the  volt- 
age to  which  they  are  to  be  subjected. 

The  capacitance  of  such  a  unit  is  its  current  for  unit  rate 
of  change  of  voltage.  Since  this  current  depends  upon  the 
number  of  electrons  in  a  cross-section  which  can  move  in 
position,  it  is  to  be  expected  that  this  capacitance  will  be 
proportional  to  the  cross-sectional  area  of  the  sheet  of  die- 
lectric. Also,  since  the  extent  to  which  the  electrons  will 
move  depends  upon  the  force  to  which  they  are  subjected,  it 
is  to  be  expected  that  the  capacitance  of  such  a  condenser 
will  be  inversely  proportional  to  the  thickness  of  the  dielectric, 
and  hence  proportional  to  the  voltage  gradient  applied  to 
the  electrons.  Such  is  found  experimentally  to  be  the  case. 
The  capacitance  is  also  proportional  to  the  dielectric 


440      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

stant,  K,  of  the  material  used.  We  may  thus  write  for 
the  capacity  of  a  parallel-plate  condenser 

'    ;    /  '  C  =  a^-,  (2) 

where  a  is  the  proportionality  factor.  If  we  take  for  con- 
venience the  dielectric  constant  of  air  to  be  unity,  then  the 
value  of  a  comes  out,  in  c.g.s.  units, 

a  =47r  X  9  X  1020' 

This  peculiar  factor  comes  in  by  reason  of  the  way  in  which 
capacitance  was  initially  defined.  We  may  therefore  write, 
in  c.g.s.  units, 

1        KA 

=  -abfarads' 


where  C  is  the  capacitance  in  abfarads,  A  is  the  cross-section 
of  the  dielectric  in  square  centimeters,  S  is  the  thickness  in 
centimeters  and  K  is  the  dielectric  constant.  In  prac- 
tical units,  we  have 


C  =  0.08842  X  lO-6-        microfarads,  (5) 

o 

where  C  is  in  microfarads,  the  other  quantities  being  as 
before. 

Prob.  8-13.  A  condenser  is  constructed  of  2000  sheets  each  of 
paraffined  paper  and  tinfoil.  The  paper  sheets  are  0.008  cm. 
thick  and  the  area  of  each  sheet  actually  between  the  tinfoil 
sheets  is  16  X  20  cm.  At  what  rate  must  the  voltage  be  varied 
across  this  condenser  to  produce  a  current  of  2.43  amperes? 

Prob.  9-13.  If  mica  sheets  0.003  cm.  thick  are  substituted 
for  the  paraffined  paper  in  the  condenser  in  Prob.  8-13,  at  what 
rate  must  the  voltage  be  varied  to  produce  2.43  amperes? 

120.  Charge  on  a  Condenser.  Electrons  repel  one  an- 
other so  powerfully  that  they  become  evenly  distributed 
throughout  the  body  of  a  metal.  They  are  also  crowded 


DIELECTRICS 


441 


out  to  the  surface,  but  cannot  leave,  except  when  the  tem- 
perature is  high,  on  account  of  the  attraction  of  the  metal 
for  them.  They  act  as  a  practically  incompressible  fluid 
in  the  body  of  the  metal.  However,  if  a  large  voltage  is 
applied,  a  few  more  can  be  crowded  into  the  surface. 

This  is  especially  true  if  a  second  metal  surface  is  close 
by  which  is  oppositely  charged,  that  is,  which  has  a  de- 
ficiency from  its  normal  supply  of  electrons,  for  the  surface 
electrons  are  attracted  by  the  metal  of  the  second  plate. 
This  partially  accounts  for  the  fact  that  two  plates  separ- 
ated by  a  vacuum  can  act  as  a  condenser,  although  it  does 
not  explain  the  experimental  fact  that  a  current  also  passes 
in  the  space  between  the  plates. 


FIG.   210. 


=HJ= 

A  hydraulic  analogy  to  a  condenser.     It  consists  of  a  pipe 
with  a  rubber  diaphragm  stretched  ..across  it. 


When  a  dielectric  is  placed  between  the  plates,  more 
electrons  can  be  crowded  into  the  metal  surface  because  of 
the  adjacent  molecules  of  insulating  material  in  which  the 
electrons  have  moved  over  in  position.  These  neighboring 
molecules  then  attract  the  electrons  of  the  metal. 

An  electric  condenser  has  an  excellent  hydraulic  analogy 
in  a  rubber  diaphragm  stretched  across  a  pipe  as  shown  in 
Fig.  210. 

If  this  pipe  is  connected  to  a  centrifugal  pump,  as  in  Fig. 
211,  and  the  pump  is  started  running,  the  diaphragm  will 
be  stretched  as  shown,  and  will  stay  stretched  as  long  as 


442      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

the  pump  operates.  During  the  period  when  the  pump  is  ac- 
celerating, there  will  be  a  flow  of  water  in  the  pipe  which  will 
soon  cease.  If  the  pump  is  stopped,  there  will  be  a  flow  of 


FIG.  211.     When  the  pump  is  started  a  current  flows  for  a  short  period 
until  the  diaphragm  is  stretched  to  its  maximum. 

water  back,  and  the  diaphragm  will  spring  back  to  its  mid- 
position. 

Similarly,  if  an  electric  condenser  is  connected  in  series 

with  a  generator,  as  shown 
in  Fig.  212,  and  the  gen- 
erator started,  there  will 
be  a  brief  flow  of  current 
around  the  circuit  for  a 
moment  after  the  generator 
is  started,  but  the  flow  will 
soon  stop  if  the  generator 
speed  and  hence  the  volt- 
age remain  constant.  If 


FIG.  212.  If  the  switch  S  is  closed 
and  the  generator  started,  a  cur- 
rent will  flow  for  a  moment  as  in 
Fig.  211. 


the  generator  is  stopped, 
and  the  voltage  thus  re- 
moved, there  will  be  a  flow  back  through  the  circuit  for  a 
short  period  of  time. 


DIELECTRICS  443 

Suppose  that  instead  of  stopping  the  generator  we  first 
open  a  switch  at  S.  This  is  equivalent  to  closing  a  valve  at 
A  in  Fig.  211.  In  such  a  case  the  current  cannot  flow  back 
even  although  the  generator  is  stopped.  We  now  say  that 
the  condenser  is  charged.  The  electrons  tend  to  spring 
back  in  the  dielectric  but  they  cannot  do  so  because  the 
electrons  in  the  plate  act  as  if  they  were  incompressible. 
There  is  a  voltage  between  the  ends  of  the  circuit  which  are 
open  which  is  equal  to  the  generator  voltage  at  the  instant 
the  switch  was  opened.  We  say  that  the  condenser  is 
charged  to  this  voltage. 

The  current  in  the  condenser  we  have  seen  to  be 


Integrating  this  expression,  we  have 

t.  (7) 


The  current  i  is  the  number  of  electrons,  measured  in  cou- 
lombs, flowing  per  second  along  the  circuit,  or  it  is  the  quan- 
tity of  electricity  per  second.  That  is, 


where  Q  is  the  quantity  of  electricity.     Insert  this  in  the 
above,  and  we  have 


or 


This  states  that  the  voltage  across  a  condenser  is  equal  to 
the  quantity  or  charge  which  has  been  forced  into  it,  di- 
vided by  its  capacitance. 


444      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 
Expressed  in  the  form 


(ID 


it  leads  to  another  definition  of  capacitance,  namely:  the 
capacitance  of  a  condenser  is  equal  to  its  charge  per  applied 
volt.  Thus  a  condenser  has  a  capacitance  of  a  microfarad 
when  an  applied  potential  of  one  volt  will  force  into  it  a 
charge  of  a  microcoulomb.  There  are  hence  two  equiva- 
lent definitions  for  capacitance,  just  as  there  are  for  induc- 
tance. 

Prob.  10-13.  A  condenser  of  12  microfarads  capacitance  is 
connected  to  a  220-volt  battery.  How  many  ampere-seconds 
of  charge  will  flow  into  the  condenser? 

Prob.  11-13.  What  capacitance  must  a  condenser  have  if 
110  volts  are  to  put  a  charge  of  0.016  coulomb  into  it? 

121.  Measurement  of  Capacitance.  There  are  two  ways 
of  measuring  the  capacitance  of  a  condenser,  depending 
upon  the  two  definitions. 


FIG.  213.     A  bridge  arrangement  for  measuring  capacitance. 

The  first  method  consists  of  applying  a  varying  voltage 
to  the  condenser  and  measuring  the  current  flowing,  or 
comparing  this  current  with  that  flowing  in  a  standard  con- 
denser under  the  same  conditions.  This  can  be  done  very 


DIELECTRICS  445 

simply  with  a  bridge  circuit  such  as  that  shown  in  Fig.  213. 
C  is  the  capacitance  of  a  standard  condenser  which  is  known. 
X  is  the  capacitance  to  be  measured.  Ri  and  R2  are  re- 
sistances. A  is  an  alternating-current  generator  which 
supplies  a  variable  voltage  to  the  bridge.  T  is  a  telephone 
receiver. 

The  resistance  R\  is  varied  until  no  sound  is  heard  in  the 
telephone.  When  this  is  done,  we  know  that  points  c  and 
d  are  at  the  same  potential  at  all  instants.  Under  these 
conditions,  currents  flow  in  X  and  C  which  are  at  all  in- 
stants proportional  to  their  capacitances,  for  they  are  sub- 
jected to  the  same  rate  of  change  of  voltage.  Since  there 
is  no  current  in  the  telephone,  these  same  currents  must 
flow  also  in  R\  and  R%.  But  the  drop  from  a  to  c  is  the  same 
as  that  from  a  to  d.  Hence  at  any  instant 

Riii  =  R&.  (12) 

We  know,  however,  that 


Combining  these  equations  gives 

R*      X 


or 

*-fc,    -  (15) 

so  that  when  R  i  and  Rz  are  known,  X  can  be  computed. 

This  method  assumes  that  there  is  no  resistance  in  the 
leads  to  the  condensers  and  no  leakage  through  the  die- 
lectric. In  the  next  section,  we  shall  study  the  effect  of  a 
resistance  in  series  with  a  condenser. 

The  above  way  of  measuring  a  capacitance  simply  com- 
pares it  with  a  capacitance  of  known  value.  The  follow- 
ing method,  which  depends  upon  the  second  definition  of 


446      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

capacitance,  allows  us  to  measure  a  condenser  without  com- 
paring it  with  another. 

If  a  steady  voltage  is  applied  to  a  condenser,  the  charge 
produced  in  the  condenser  divided  by  the  applied  voltage 
gives  the  capacitance  of  the  unit.  The  charge  is  equal 
to  the  total  quantity  of  electricity  passing  through  the  leads 
during  the  process  of  charging. 

We  have  seen,  in  measuring  flux,  that  the  deflection  of  a 
ballistic  galvanometer  is  proportional  to  the  quantity  of 
electricity  which  is  suddenly  passed  through  it.  Let  us 
therefore  connect  a  ballistic  galvanometer  in  series  with  the 

capacitance  to  be  measured, 
and  suddenly  close  a  switch 
to  apply  a  known  voltage  E. 
The  arrangement  is  shown 
in  Fig.  214.  Let  d  be  the 
deflection  of  the  galvano- 
meter when  the  switch  is 


FIG.  214.  A  ballistic.  galvanometer  closed.  Let  K  be  the  gal- 
method  of  measuring  capacitance,  vanometer  constant,  that 

is.    the    deflection   for   one 

microcoulomb.  Then,  since  a  deflection  d  =  K  when  one 
volt  is  applied  to  a  capacitance  of  one  microfarad,  we 
have  in  general 

X  =  ^=  microfarads.  (16) 


The  constant  K  may  be  found  in  several  ways.  One  is  to 
use  a  standard  condenser,  measure  the  deflection,  and  use 

,  K  =  ^E,  (17) 

where  d\  is  the  deflection  obtained  with  voltage  E  and  C 
is  the  known  capacitance.  This  involves  comparison,  how- 
ever, and  we  may  have  no  standard  capacitance  available. 
Another  method  is  to  calibrate  by  using  an  inductance  as 
outlined  under  flux  measurement.  A  third  way  is  to  apply 


DIELECTRICS 


447 


a  low  voltage  through  a  high  non-inductive  resistance  to 
the  galvanometer  for  a  short  instant  by  means  of  a  contact- 
making  device  such  as  a  commutator.  An  arrangement  for 
doing  this  is  shown  in  Fig.  215. 

The  contact  device  here  shown  is  a  pendulum  which  makes 
a  contact  for  an  instant  at  the  lower  part  of  its  swing.  From 
the  period  and  amplitude 
of  oscillation  of  the  pen- 
dulum and  the  length  of 
the  contact  arm,  we  can 
compute  the  short  inter- 
val of  time  Jo  during  which 
the  pendulum  causes  con- 
tact to  be  made.  The 
switch  S  is  closed  during 
one  swing  of  the  pendu- 


lum only,  and  the  deflec- 
tion of  the  galvanometer 
noted.  During  a  time  to 
there  is  a  current  through 
the  galvanometer  of  value 


FIG.  215.  The  time  of  contact  made 
by  the  pendulum  can  be  determined. 
By  means  of  this  and  the  known 
values  of  E  and  R,  the  value  of  the 
constant  of  the  galvanometer  can  be 
computed. 


E 

-^  amperes. 

ri 


The  total  quantity  passing  is  hence 

*       Wf 

Q  =  it0  =  —  coulombs, 
rt 


(18) 


(19) 


If  the  galvanometer  deflection  due  to  this  quantity  is  d", 
the  constant  K  of  the  deflection  per  microcoulomb  will  be 

d"  _  (2Q) 


or 


(21) 


This  method  of  calibrating  the  galvanometer  must  be  used 
with  considerable  care  to  obtain  accurate  results. 


448      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

When  a  condenser  is  discharged,  the  same  quantity  flows 
out  as  flowed  in  during  charge,  if  no  electrons  are  lost  by 
leakage.  Accordingly,  if  we  wish  we  may  connect  the  cir- 
cuit as  shown  in  Fig.  216  and  measure  the  quantity  passing 

the  galvanometer  during  dis- 

•\  • j         charge  of  the  condenser,  by 

throwing  the  two-way  switch 
first  to  the  battery  side  and 
then  to  the  galvanometer. 
The  equations  are,  of  course, 


n 


FIG.  216.     The  galvanometer  will     the  same  as  before, 
measure  the  quantity  of  electric- 
ity passed  through  it  upon  dis-         Prob.  12-13.      In   a   bridge 
charge  of  the  condenser.  arranged   as   in   Fig.   213,   no 

sound  is  heard  in  the  telephone 

receiver  when  R^  =  1.297  ohms  and  R2  =  3.000  ohms.     What 

is  the    capacitance   of  X  if   the    capacitance    of    C   is    2.461 

microfarads? 

Prob.  13-13.  In  Fig.  213,  suppose  that  A  and  T  are  inter- 
changed in  position.  What  will  be  the  equation  of  the  bridge 
for  this  condition? 

Prob.  14-13.  A  pressure  of  0.100  volt  is  applied  for  0.010 
second  to  a  ballistic  galvanometer  having  a  total  resistance  of 
50,000  ohms.  The  galvanometer  deflects  2.00  cm.  A  con- 
denser freshly  charged  to  20  volts  is  discharged  through  the 
ballistic  galvanometer  and  it  deflects  15.99  cm.  What  is  the 
capacitance  of  the  condenser? 

Prob.  15-13.  What  capacitance  must  a  condenser  have  if, 
when  charged  to  100  volts,  it  will  deflect  the  galvanometer 
of  Prob.  14-13,  20  cm.  on  being  discharged  through  it? 

122.  Charging  a  Condenser  Through  a  Resistance.  Let 
us  examine  the  transient  which  occurs  in  charging  a  con- 
denser through  a  resistance.  That  is,  let  us  study  the 
manner  in  which  the  current  varies  in  the  circuit  of  Fig. 
217  during  the  interval  after  the  switch  is  closed. 

Call  the  current  in  the  circuit  at  any  instant  i.  No 
matter  how  this  current  varies,  the  total  quantity  which 


DIELECTRICS  449 

passes  through  the  circuit  up  to  any  given  time,  t,  can  be 
written 

q=f'idt.  (22) 

«/0 

This  quantity  is  the  charge  on  the  condenser  at  this  instant. 
The  voltage  to  which  the  con- 
denser is  charged  is  then 

3  =  I  I    r  Jf/.    (23)       _L 


Now  there  are  two  voltages 
in    the    circuit,    the    applied 

voltage  of  the  battery  and  the     Fla  ?"•    A  cj,rcuit  c°ntainin*5 

resistance  and  capacitance  m 
back  voltage  to  which  the  con-        series-. 

denser  is  charged.     The  dif- 
ference between  these  two  acts  to  produce  a  current  through 
the  circuit.     That  is, 

E  -i  Cidt 


-  i  f'i 


Rewrite  this  in  the  form 

E-iR=±£idt,  (25) 

and  take  the  derivative  with  respect  to  t  of  both  sides  of  the 
equation.     Note  that  by  definition 


t  =  •*•  (26) 

The  differential  equation  of  the  circuit  is  therefore 


To  solve,  separate  the  variables 

di  * 

J--RC' 


450      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 
and  integrate 

XtC 

where  A  is  a  constant  of  integration.  To  determine  A, 
note  that  when  t  =  0,  that  is,  when  the  switch  is  first  closed, 
the  current  is  simply  E/R,  for  at  that  time  there  is  no  charge 
on  the  condenser  to  oppose  the  flow  of  current.  This  relation 
may  be  obtained  also  from  equation  (19)  by  inserting  t  =  0. 
Since 

t  =  0 
and  .  _  E 

we  have 

;>  (30) 


and  inserting  this  in  (29) 

,  (31) 


R 
This  equation  can  also  be  written  in  the  form 

jjj  =<r*c>  (33) 

R 
or 

*"lr""  (34) 

This  is  the  equation  for  the  charge  of  a  condenser  through 
a  resistance.  It  is  plotted  in  Fig.  218.  We  note  that  when 
the  switch  is  closed,  the  current  starts  off  at  full  value  E/R 
and  exponentially  falls  to  zero.  This  curve  is  exactly  similar 
to  the  transient  by  which  the  current  dies  out  in  an  inductive 


DIELECTRICS 


451 


circuit.  In  the  same  way  as  there  treated,  we  may  note 
that  when  t  =  RC,  the  current  will  have  fallen  to  1/e  of 
its  initial  value.  The  quantity  RC  is  hence  called  the  time 
constant  of  the  condenser  circuit. 


FIG.  218.     The  charging  current  of  a  condenser.     The  time  repre- 
sented by  RC  is  called  the  time  constant  of  the  circuit. 


The  rate  of  decrease  of  the  current  at  any  instant  is  given 


by 


di 
dt 


~ 


(35) 


obtained   directly   from    (34).     Inserting   t  =  0,   we   have 

(£)...-&  (36) 

as  the  initial  rate  of  decrease  of  the  current,  If  the  current 
had  continued  to  decrease  at  this  rate  for  a  time  RC,  it- 
would  have  fallen  through  the  value 


R  ' 


(37) 


that  is,  it  would  have  fallen  to  zero. 

The  time  constant  hence  may  also  be  expressed  as  the 


452      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

time  it  would  take  the  current  to  fall  to  zero  if  it  continued 
to  decrease  at  its  initial  rate. 

The  charge  on  the  condenser  at  any  instant  is  given  by 
integrating  the  expression  for  current,  or 


A 

Jo 


dt.  (38) 

o 

This  gives 

(39) 


"I 


-  ECe 

lo 

=  EC  (i  -rsb).  (40) 

The  final  charge  on  the  condenser  is 

Q=EC, 
so  that  we  may  write 

q  =  Q(l-c-^).  (41) 

The  charge  therefore  increases  exponentially  to  its 
final  value  in  accordance  with  the  curve  of  Fig.  219.  The 
same  time  constant  as  before  applies  to  this  curve  also. 

Prob.  16-13.  The  circuit  of  Fig.  217  has  the  following  con- 
stants : 

R  =  50  ohms, 
C  =  5  microfarads, 
E  =  100  volts, 

r    =  2  ohms  (internal  resistance  of  the  battery), 
(a)  At  what  rate  will  the  current  be  decreasing  0.006  second 
after  the  switch  is  closed? 

(6)  What  time  will  be  necessary  for  the  current  to  decrease 
to  half  its  initial  value? 

(c)  What  is  the  initial  rate  of  decrease? 

(d)  What  is  the  value  of  the  current  when  t  =  CR1 


DIELECTRICS 


453 


Prob.  17-13.     In  the  circuit  of  Fig.  217: 

(a)  What  will  be  the  charge  on  the  condenser  0.006  second 
after  the  switch  is  closed? 

(6)  What  time  will  it  take  to  put  half  the  total  charge  on  the 
condenser? 

(c)  What  is  the  value  of  the  charge  when  t  =  CRt 


-RC- 

FIG.  219.  The  charge  on  a  condenser  increases  in  accordance  with 
this  curve.  The  time  represented  by  RC  is  called  the  time  con- 
stant of  the  circuit. 

Prob.  18-13.  After  the  condenser  in  the  circuit  of  Fig.  217 
has  become  fully  charged,  a  second  battery  of  negligible  internal 
resistance  and  giving  75  volts  terminal  voltage  is  suddenly  added 
to  the  circuit,  in  series  with  the  one  already  acting. 

(a)  At  what  rate  will  the  current  be  decreasing  0.002  second 
after  the  second  battery  has  been  added? 

(#)  What  time  must  elapse  before  the  current  reaches  half 
the  value  it  has  at  the  instant  of  adding  the  second  battery? 

(c)  What  time  does  it  take  to  put  in  half  the  final  added 
charge? 

(d)  What  is  the  value  of  the  added  charge  when  the  time 
since  adding  the  second  battery  has  the  value  CR1 

Prob.  19-13.  Determine  from  measurements  made  on  the  curve 
of  the  charging  current  in  Fig.  220,  the  capacitance  of  the  conden- 
ser used. 

123.  Discharge.  The  equation  for  discharge  follows 
directly  from  that  of  charge.  In  the  circuit  of  Fig.  221, 


454      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

we  will  assume  the  condenser  initially  charged  to  a  potential 
E,  and  then  the  switch  closed  at  time  t  =  0. 

The  voltage  of  the  condenser  then  is  at  any  instant  there- 
after 

(42) 


FIG.  220.  An  oscillogram  of  the  charging  current  of  a  condenser  with 
a  1000-ohm  resistor  in  series  when  115  volts  are  applied  to  the  com- 
bination. The  wave  at  the  bottom  is  a  60-cycle  wave  from  which 
the  time  may  be  obtained.  Taken  by  Prof.  F.  S.  Dellenbaugh  in 
the  Research  Laboratories  of  the  Electrical  Engineering  Dept.,  Mass. 
Inst.  of  Technology. 


n 


FIG.  221.     A  circuit  containing  capacitance  and  resistance. 

and  since  this  is  the  only  voltage  acting  in  the  circuit,  the 
current  can  be  obtained  by  dividing  by  R 


(43) 


R 


DIELECTRICS 
Rewriting  this  in  the  form 


455 


(44) 


it  is  seen  to  be  exactly  the  same  equation  as  (25)  for  charge, 
except  that  E  now  means  an  initial  voltage  instead  of  a 
battery  voltage.  The  solution  is  accordingly  exactly  the 
same, 

f'-fe-ES  (45) 


FIG.  222.     Oscillogram  of  the  discharge  current  of  the  condenser  of 
which  Fig.  220  is  the  charging  current.     Pro/.  F.  S.  Dellenbaugh. 

The  curves  of  current  upon  charge  and  discharge  are 
hence  identical.  This  fact  is  illustrated  by  Fig.  220  and  222 
which  are  oscillograms  of  the  charge  and  discharge  currents 
of  the  same  condenser. 

The  quantity  on  the  condenser  thus  decreases  during 
discharge  at  the  same  rate  as  it  increased  during  charge. 
Hence  the  curve  of  q  is  that  of  Fig.  223,  and  the  equation 


t 

RC  . 


(46) 


Prob.  20-13.  The  constants  for  the  circuit  of  Fig.  221  are 
C  =  6.28  mf.  and  R  =  18  ohms.  The  condenser  is  charged 
to  a  potential  of  150  volts  and  the  switch  is  closed.  Compute 


456      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


(a)  Current  at  instant  of  switch  closing, 
(6)  Rate  of  decrease  of  current  at  instant  of  switch  closing, 
(c)  Time  in  which  charge  on  the  condenser  is  reduced  one-half, 
(rf)  Charge  left  on  condenser  after  0,008  second. 


FIG.  223.  The  discharge  curve  showing  the  relation  between  the  charge 
on  a  condenser  and  the  time  of  discharge.  The  charge  on  the  con- 
denser decreases  upon  discharge  through  a  resistance  according 
to  this  curve. 

124.   Energy  Relations.     Referring  to  Fig.  217,  the  equa- 
tion 

E  =  iR  +  ^   /  idt  (47) 

applies  to  the  current  and  voltage  relations  during  charge  and 
expresses  the  fact  that  at  any  instant  the  applied  voltage 
E  is  partly  used  in  overcoming  the  iR  drop  in  the  resistance, 

and  partly  in  opposing  the  back  voltage,  ~  /  i  dt,  of  the 
condenser.  Multiply  the  equation  by  i,  and  we  have 


Ei 


j/<* 


(48) 


This  expresses  the  fact  that  of  the  power  input  Ei  at  any 
instant,  a  part  i2R  is  used  in  heating  the  resistance,  and  the 
remainder 


is  stored  in  the  condenser. 


1  I  idt 


DIELECTRICS  457 

When  the  current  is  varying  in  accordance  with 

(49) 


the  rate  of  energy  storage  in  the  condenser  becomes,  upon 
substituting  this  value  of  i, 

E        * 


E2  --L  -L 

—  TT€   RC  (—  e    RC), 
ft 

E2      2t 
=  --e~RC.  (51) 

The  total  energy  stored  is  the  integral  of  this  expression  over 
the  total  time  used  in  charging  the  condenser:  that  is, 

/»°°      E2      2t 

Wc=  /      -^e-Rcdt.  (52) 

Jo          K 

The  integration  gives 

JpZC1          2t  -i°° 

Wc  =  ^t-Rc]Q  (53) 

or 

Wc  =  ^-  joules.  (54) 

This  is  the  expression  for  the  energy  stored  in  a  condenser 
when  it  is  charged  to  a  potential  E.  It  will  be  remembered 
that  when  a  current  /  is  flowing  through  an  inductance  L, 
the  energy  storage  was 


WL  =         joules.  (55) 

A 

The  two  expressions  are  seen  to  be  very  similar.  The 
energy  stored  in  a  condenser  is  analogous  to  a  potential 
energy,  while  that  stored  in  an  inductance  is  similar  to  a 
kinetic  energy. 

It  is  interesting  to  note  the  energy  lost   in   the  series 


458     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

resistance  during  the  process  of  charging  the   condenser. 
This  is, 

WB=  C™i2Rdt.  (56) 

Jo 

Putting  in 

E  __L 

i  =  -  c  RC  ,  (57) 

ti 

we  have 

r*  E2     2t 

WR=J      j^racM,  (58) 

and  integrating 

IT,-  -££.-£]"  (59) 

'  <m 

This  gives  us  a  curious  fact.  When  a  condenser  is  charged 
through  a  resistance  from  a  constant  potential  source,  ex- 
actly half  of  the  energy  put  in  is  lost  in  the  resistance,  and 
the  other  half  is  stored  in  the  condenser.  This  is  true  no 
matter  whether  the  resistance  is  large  or  small.  If  the 
condenser  is  charged  from  a  voltage  source  which  is  gradu- 
ally increasing  as  the  condenser  charges,  not  nearly  as  much 
energy  will  be  lost. 

Prob.  21-13.  A  condenser  of  1.00  mf.  capacitance  is  charged 
to  500  volts.  How  much  energy  is  stored  in  it?  Express 
answer  in  ergs,  joules  and  kilowatt-hours. 

Prob.  22-13.  How  much  energy  is  stored  in  a  condenser  of 
0.005  mf.  capacitance  when  charged  to  a  pressure  of  100,000 
volts?  Express  answer  in  ergs,  joules  and  kilowatt-hours. 

Prob.  23-13.  The  condenser  of  Prob.  21-13  has  a  resistance 
of  25  ohms  in  series  with  it. 

(a)  At  what  rate  is  energy  being  stored  in  the  condenser  0.08 
second  after  the  switch  is  thrown? 

(6)  At  what  rate  is  energy  being  consumed  in  the  resistance 
at  the  same  instant? 


DIELECTRICS 


459 


125.  Mechanical  Force  on  a  Condenser.  When  two 
bodies  are  charged,  there  are  always  small  mechanical 
forces  acting  between  them,  called  electrostatic  forces. 
These  forces  tend  to  pull  the  bodies  together  when  the 
charges  are  opposite  in  sign, 
or  to  push  them  apart  if  the 
charges  are  similar.  Electro- 
static attraction  is  somewhat 
similar  to  magnetic  attraction, 
but  is  not  as  important  prac- 
tically. 

The  consideration  of  the 
last  section  will  enable  us  to 
compute  the  force  acting 
between  the  plates  of  a 
parallel  -  plate  condenser 
when  the  condenser  is  charged  to  a  difference  of  poten- 
tial E. 

Consider  the  condenser  of  Fig.  224.    Assume  that  we 
have  charged  it  to  a  potential  E  and  then  disconnected  it 

from  the  circuit.  The 
energy  stored  in  the 
condenser  is 


FIG.  224.  A  condenser  the  two 
plates  of  which  are  separated 
by  a  distance  S. 


=       -  joules.  (60) 


Consider     one     plate 
attached   fast.      Let  us 

apply  a  force  F  to  the 
FIG.  225.     The  plates  of  the  condenser         ,,  ,   ,  ,         ,,    ., 

in  Fig.  224  have  been  here  pulled  °ther  Plate  and  Pul1  .rt 
apart  by  a  force  F  to  a  distance  away  an  additional  dis- 
S  +  dS.  tance  dS  as  shown  in 

Fig.   225.     In  so  doing 
we  have  expended   work  to  the  amount 


dW  =  FdS, 


(61) 


460      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

which,  assuming  no  friction  or  other  losses,  must  appear  as 
additional  energy  stored  in  the  condenser.  The  energy 
now  stored  in  the  condenser  is 

TF,  =  C-^,  (62) 


where  C\  and  E\  are  the  new  capacitance  and  voltage  re- 
spectively. 

For  the  new  separation  the  capacity  is  decreased  to  the 
value 


since  the  capacity  of  a  parallel-plate  condenser  is  inversely 
proportional  to  the  separation  of  the  plates. 

Also,  since  the  number  of  electrons  on  the  plates  has  re- 
mained unchanged  during  the  process,  and  since  the  po- 
tential on  a  condenser  is  the  charge  divided  by  the  capacity, 
we  have 

El=*+E.  (64) 


Inserting  these  two  expressions,  we  then  have 

^±^E)*  (65) 


=  ^W,  (66) 

and  hence 

dW  =  Wi  -  W  (67) 


-f?- 

But  we  had  above 

dW  =  FdS.  (69) 


DIELECTRICS  461 

So,  equating  these  expressions, 

/  =:  ^  x  107  dynes.  (70) 

This  expression  gives  the  force  acting  on  the  plates  of  a 
charged  parallel-plate  condenser.  If  all  quantities  on  the 
right  are  in  the  c.g.s.  system,  the  force  will  be  given  in 
dynes  and  the  factor  107  will  not  be  needed. 

We  can  put  the  expression  in  a  different  form  by  inserting 
the  expression  for  the  capacitance, 

C  =  36710*  ^Tabfarads-  <71> 

so  that 

1         KAE* 

J     "~~"     Wffc        -t  /"\on  do          vlV HCD •  \t£jj 

72  TrlO20       o2 

The  potential  gradient  between  the  plates  or  the  electro- 
motive force  per  unit  of  length  is  uniform  and.  of  value 

F=|-  (73) 

The  electric  potential  gradient  F  is  exactly  analogous  to  the 
magnetic  potential  gradient  H,  sometimes  called  the  mag- 
netizing force.  It  will  be  remembered  that  in  a  uniform 
magnetic  field 

H  =  y  >  (73a) 

where 

H  =  the  magnetic  potential  gradient, 
y  =  the  magnetomotive  force, 
I  =  the  length  of  the  magnetic  flux  path. 

Equation  (73a)  for  a  magnetic  field  is  exactly  analogous  to 
equation  (73)  for  an  electrostatic  field.  The  magnetomotive 
force  y  is  ordinarily  measured  in  ampere-turns  or  in  gilberts 
and  the  length  I  of  the  field  in  centimeters.  The  magnetic 


462      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

potential  gradient  is  therefore  in  units  of  ampere-turns  per 
centimeter,  or  in  gilberts  per  centimeter.  1  ampere-turn 
=  0.4  TT  gilberts. 

Similarly  we  measure  the  electromotive  force  E  in  volts 
or  in  statvolts*  and  the  length  of  the  field  S  in  centimeters. 
1  statvolt  =  300  volts.  Thus  the  electric  potential  gradient  is 
in  units  of  volts  per  centimeter  or  in  statvolts  per  centimeter. 

The  analogy  extends  still  further.  The  magnetic  field 
is  thought  of  as  containing  lines  of  magnetic  force.  Similarly 
the  electric  field  is  thought  of  as  containing  lines  of  electro- 
static force.  In  the  case  of  the  magnetic  field,  we  mul- 
tiply the  magnetic  potential  gradient  H  at  a  point  in  the 
field  by  the  permeability  n  of  the  medium  at  that  point,  and 
obtain  the  magnetic  flux  density  B  at  that  point.  So  in  the 
case  of  an  electrostatic  field,  we  multiply  the  electric  poten- 
tial gradient  F  at  a  given  point  in  the  field  by  the  per- 
mittivity K,  (or  dielectric  constant)  of  the  medium  at  that 
point,  and  obtain  the  electrostatic  flux  density  D  at  that 
point.  Expressed  as  equations, 

B  =  ?H,  (74) 

D  =  KF.  (74a) 

Of  course,  the  proper  units  for  D,  K  and  F  must  be  used 
to  make  this  equation  numerically  correct.  We  will  take 
up  these  units  below. 

If  now  the  value  of  F  as  found  in  equation  (73)  is  substituted 
in  equation  (74a),  we  obtain 


~s" 

In  terms  of  K,  equation  (72)  becomes 
]_ _  AD*d  neg 

which  may  be  written 

/  =  ^>no*xSdynes-        (75a) 

*  A  statvolt  per  centimeter  is  the  potential  gradient  which  will  pro- 
duce unit  electrostatic  flux  density  in  a  medium  of  unit  permittivity. 


DIELECTRICS  463 

This  equation  is  again  analogous  to  the  equation  for  the 
magnetic  pull 

D2  A 

(756) 


the  factor  0  coming  into  equation  (75a)  because  of 

the  arbitrary  choice  of  unity  for  the  dielectric  constant  of  air. 

The  electrostatic  pull  is  thus  proportional  to  the  cross- 
sectional  area  and  to  the  square  of  the  electrostatic  flux 
density,  just  as  the  magnetic  pull  is  proportional  to  the 
cross-sectional  area  and  to  the  square  of  the  magnetic  flux 
density.  In  addition  the  electrostatic  pull  is  inversely 
proportional  to  the  dielectric  constant  of  the  medium. 
We  always  computed  the  magnetic  pull  in  air,  but  in  any 
other  medium  it  would  similarly  have  been  found  inversely 
proportional  to  the  permeability. 

The  mechanical  force  due  to  the  electrostatic  field  is 
usually  small.  For  example,  suppose  we  have  two  plates 
each  10  centimeters  square,  separated  1  centimeter  and  sub- 
jected to  an  electromotive  force  of  100  volts  or  1010  abvolts. 
Suppose  they  are  separated  by  air,  of  dielectric  constant 
unity.  Then  inserting  the  values  in  the  formula  above, 


i  (76) 


72  TrlO20 
100 


727T 

,11  f 
2000 


about  \  dyne, 


=  about  ^^r  gram,  (77) 


a  force  so  small  that  it  could  be  detected  with  difficulty. 

However,  suppose  the  plates  were  in  a  hard  vacuum  so 
that  the  potential  between  them  could  be  raised  to  100,000 
volts  without  flash-over.  Since  the  force  is  as  the  square  of 
the  applied  voltage,  it  is  now  (1000)2  times  as  great  as  before, 
that  is,  about  500  grams,  or  a  little  more  than  a  pound. 


464      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


Electrostatic  attraction  can  thus  be  made  use  of  in  meters 
for  measuring  high  voltage,  by  causing  the  attraction  to  move 
a  delicately  suspended  vane,  as  shown  in  Fig.  226  and  227. 

On  the  other  hand,  the 
force  must  be  carefully 
balanced  in  a  high  voltage 
thermionic  rectifier,  for  the 
very  high  potential  gradient 
met  with  in  such  a  vacuum 
device  may  pull  over  or 
even  break  the  hot  fil- 
ament. 

Prob.  24-13.  In  an  electro- 
static voltmeter  of  the  form 
shown  in  Fig.  226,  the  effect- 
ive area  of  each  plate  is 
80  square  centimeters.  The 
distance  is  8  centimeters. 
What  force  acts  on  the 
plates  when  a  potential  dif- 
ference of  25,000  volts  exists 
between  them? 

Prob.  26-13.  In  designing 
an  electrostatic  voltmeter  of 
the  type  shown  in  Fig.  226, 
for  80,000  volts,  it  is  desired 
to  produce  an  attractive  force 
°f  tV  gram.  Determine  the 
other  dimensions  of  the  volt- 
meter, using  two  plates 
of  the  same  size  and  having 
a  factor  of  safety  of  4  against 
flash-over.  Neglect  fringing 
force. 


FIG.  226.  An  electrostatic  volt- 
meter for  measuring  high  volt- 
ages. The  electrostatic  force 
between  the  pans  is  a  function 
of  the  voltage  between  them. 
General  Electric  Co. 

of    the    electrostatic    lines    of 


Prob.  26-13.  A  certain  condenser  consists  of  two  metal 
plates  each  having  an  area  of  150  square  centimeters  and  placed 
0.60  centimeter  apart.  The  dielectric  is  air.  The  plates  are 
charged  to  a  difference  of  potential  of  500  volts  and  the  source 
of  the  potential  removed. 


DIELECTRICS  465 

(a)  How  much  work  would  be  done  in  separating  the  plates 
until  they  were  1.40  centimeters  apart? 

(6)  What  would  be  the  difference  of  potential  between  the 
plsrtes  if  a  mica  sheet  0.4  centimeter  thick  were  now  thrust 
between  them? 


FIG.  227.     Another  type  of  electrostatic  voltmeter. 
General  Electric  Co. 

126.  Electrostatic  Fields.  In  the  previous  article  we 
have  met  the  idea  of  the  electrostatic  flux  density  D.  We 
can,  in  fact,  plot  the  electrostatic  lines  of  force  just  as  we  have 
previously  plotted  magnetic  lines.  Thus  in  Fig.  228  is 
plotted  the  electrostatic  field  about  a  pair  of  parallel  wires  in 
air  between  which  there  exists  an  electromotive  force.  The 
density  of  the  electrostatic  flux  lines  D  at  any  point  in  air 
is  equal  to  the  potential  gradient  at  that  point.  If  at  any 
point  this  gradient  is  greater  than  the  dielectric  strength  of 
the  insulation  (the  air  in  this  case)  there  will  be  break  down, 
and  corona  or  spark. 

For  a  parallel  plate  condenser  we  had 


D  =  3  x      io1"1^  Per  scLuare  centimeter,          (78) 


466      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 
but  for  such  a  condenser  we  had,  in  equation  (71), 

(79) 


where  a  is  a  proportionality  factor  equal  to  20  '  anc*  so 


FIG.   228.     The  electrostatic  field  between  a  pair  of  charged 
parallel  conductors  of  round  cross-section. 

Inserting  this  value  in  equation  (78),  we  have 

^nes  Per  sQuare  centimeter.       (81) 


1  OKA 

X  lU   ) 

This  expression  states  that  the  electrostatic  flux  density  is 
proportional  to  the  charge  per  unit  of  area  of  the  plates, 


DIELECTRICS  467 

that  is,  to  the  number  of  excess  electrons  per  square  cen- 
timeter of  plate. 

In  other  words,  in  mapping  the  electrostatic  field  for  a 
parallel-plate  condenser,  we  should  draw  one  electrostatic 
line  to  each  charge  consisting  of  a  certain  definite  number  of 
electrons. 

Since  in  c.g.s.  units 

'     ':  (82) 


it  follows  that  we  should  draw  one  electrostatic  line  to  each 

3  X  1010  1 

4^(3  X  10'T  =  4T(8  X  10")  abcoulomb  of  charge-     («) 

That  is,  we  should  draw  4  7r(3  X  1010)  lines  to  each  abcou- 
lomb. 

The  quantity  of  electricity 

(3  X  1010)    abcoulombs  (84) 

is  called  a  stat  coulomb.  It  is  the  quantity  which  situated 
at  1  centimeter  distance  from  a  similar  quantity  will  repel 
it  with  a  force  of  1  dyne. 

4  TT  electrostatic  lines  thus  proceed  from  each  statcoulomb 
of  charge.  It  was  the  early  choice  of  this  value  which  led 
to  the  presence  of  TT  in  the  constants  above.  The  quantity 

3  X  1010  which  appears  in  the  relation  between  the  units  is 
equal,  in  centimeters  per  second,  to  the  velocity  of  light. 
The   numerical  relations   which  exist  among  the   various 
units  of  charge  are  as  follows: 

1  faraday  =  9,654  abcoulombs, 

1  abcoulomb     =  10  coulombs, 
1  coulomb         =  3  X  109  statcoulombs, 
1  statcoulomb  =  2.1  X  109  electrons. 

When  we  map  any  electrostatic  field  we  should  consider 

4  TT  lines  to  proceed  from  each  statcoulomb  of  charge  of  one 
sign,  and  terminate  on  an  equal  charge  of  opposite  sign. 


468      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


The  flux  density  at  any  point  will  then  be  equal  to  the 
voltage  gradient  in  abvolts  per  centimeter.  Of  course  in 
drawing  an  electrostatic  field  sectionally,  it  is  usually  con- 
venient to  actually  plot  in  only  a  very  small  fraction  of 
the  lines  as  above  obtained. 

127.  Capacitance  of  a  Pair  of  Long  Parallel  Aerial  Con- 
ductors. In  developing  the  equation  for  the  inductance 
of  a  pair  of  parallel  aerial  conductors,  we  first  found  the 
value  of  the  magnetizing  force  or  magnetic  potential  gradient 
H  at  any  point  in  the  space  surrounding  a  long  straight 
conductor.  Similarly  the  capacitance  of  a  pair  of  aerial 

conductors  can  be  found  by 
first  finding  the  electric 
potential  gradient  F  at  any 
point  in  the  air  around  such 
a  conductor. 

Fig.  229  is  constructed 
similar  to  Fig.  85  which  was 
used  to  find  the  inductance 
of  a  conductor.  Let  Q  in 


\dFm 

FIG.  229.  The  force  dF  is  pro- 
duced by  the  electric  charge  on 
the  element  dx  of  the  con- 
ductor. 


statcoulombs  represent  the 
charge  per  centimeter  of 
length  on  the  long  conductor. 
The  charge  on  dx  length 
will  then  be  Qdx. 

It  will  be  remembered  that  the  current  flowing  through 
an  element  dx  of  a  straight  conductor  produces  a  magnetizing 
force  or  magnetic  potential  gradient  H  at  a  point  P  outside 
the  conductor,  the  value  of  which  is  found  by  the  equation 


Idx 
dH  =  -™-cos  0. 


(85) 


Similarly  the  strength  of  the  electrostatic  force  dF,  Fig.  229, 
or  electric  potential  gradient  in  this  direction  at  the  point  P, 
due  to  a  charge  Qdx,  is  expressed  by  the  equation 

s8,  (86) 


DIELECTRICS  469 

where 

dF  is  the  component  of  electrostatic  force  or   potential 

gradient  at  P  perpendicular  to  the  conductor:    it  may 

be  measured  in  dynes  on  a  unit  charge  placed  at  the 

point  or  in  stat volts  per  centimeter; 

I  is  the  distance  in  centimeters  of  point  P  from  the  element 

dx\ 
Q  is  the  charge  on  the  conductor  in  statcoulombs  per 

centimeter; 

6  is  the  angle  between  the  electrostatic  force  at  P  due  to 
dx  and  the  component  of  this  force  perpendicular  to  the 
conductor. 

The  electrostatic  force  or  potential  gradient  F  at  P  due  to 
the  charge  on  a  wire  (of  practically  infinite  length  in  com- 
parison with  r)  is  expressed  by  the  equation 


(87) 

But  from  Fig.  229  it  is  seen  that 

x  =  r  tan  0,  (88) 

dx  =  r  sec20  d0,  (89) 

I  =  -^--  (90) 

cos  0 

If  x  is  indefinitely  increased,  0  will  have^  —  »  and  +  »  as 

limits.     Therefore  substituting  these  limits  and   (88),  (89) 
and  (90)  in  (87)  we  have 

•'    <™^ede  (91) 


(92) 
J~\       f 

Integrating, 


470      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Note  that  the  expression  for  electrostatic  force  F  in  stat- 
volts  per  centimeter  is  again  exactly  analogous  to  the  ex- 

9  T 

pression  -  -  in  gilberts  per  centimeter  for  the  magnetizing 

force  H  at  a  point  outside  of  a  conductor  carrying  a  current. 
To  find  the  capacitance  of  a  pair  of  long  parallel  aerial 
conductors,  let  A  and  B,  Fig.  230,  be  a  pair  of  such  con- 
ductors spaced  S  centi- 

\£)  V|y      having   a   radius   of    r 

centimeters    and    each 
charged    with   Q    stat- 

coulombs  per  centi- 
FIG.  230.  A  and  B  represent  two  long  meter  j  th  ^  ft 

parallel     conductors     having     unlike          .  „    -^  u 

i    ,  .     ,  voltage  of  E  statvolts 

electric  charges. 

between  them.  Con- 
sider the  charge  on  A  positive  and  that  on  B  negative. 
These  conductors  must  be  far  enough  apart  for  the  charge 
on  each  to  be  uniformly  distributed  around  each,  in  order 
that  the  following  development  be  valid.  In  other  words, 
S  must  be  large  in  comparison  with  r. 

The  value  of  F,  the  potential  gradient  or  electrostatic 
force,  at  any  point  P  a  distance  of  x  from  A  on  a  line  join- 
ing the  centers  of  the  conductors  A  and  B  can  be  found  as 
follows. 

The  electrostatic  force  FA  at  P,  due  to  the  charge  Q  on  A, 
is  expressed  by  the  equation 

FA=^.  (94) 

The  electrostatic  force  FB  at  P,  due  to  the  negative  charge 
—  Q  on  the  conductor  5,  is  in  the  same  direction  as  DA  and 
is  represented  by  the  expression 


H- 

o  —  X 


DIELECTRICS  471 

The  total  electrostatic  force  at  P  is  expressed  by 

F  =  FA  +  FB  =  2Q(-  +  -^—]  statvolts  per  centimeter. 

(96) 

Since  the  voltage  between  the  wires  must  be  the  integral  of 
the  potential  gradient  between  the  wires, 

/S-r 
Fdx  statvolts  (97) 

or 


S-r 


Integrating  this  expression  we  obtain 

E  =  4  Q  loge  ^—^  statvolts  .  (98) 


r 
But 

C=|-  (99) 

Therefore  substituting  (98)  in  (99)  we  have 

C  =-        ^     _  * S-r  statfarads     (10°) 

4  Q  loge  —  4  loge  - 

per  centimeter  of  line. 

1  statfarad  =  ^-r—^i  microfarad 


Therefore 

2.54  X  12  X  5280 


C  = 


2.3  X  4  X  9  X  105  log 


0  0194 

microfarads  per  mile  of  line.        (101) 


i  ~  r 

log — - — 


472      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

This  value  is  an  approximation  only  and  can  be  used  only 
when  S  is  very  large  in  comparison  with  r.  For  this  reason 

0  0194 

— =  is  usually  about  as  close  an  approximation.     In  using 

loS- 

this  equation  the  effect  of  the  ground  is  neglected,  which 
may  be  done  when  the  line  is  strung  at  considerable  height 
above  the  ground.  Even  then  trees  along  the  line  are 
likely  to  affect  the  capacitance  of  the  conductor. 

Prob.  27-13.  The  Sierra  and  San  Francisco  Power  Co. 
transmits  power  from  Stanislaus  to  San  Francisco,  a  distance  of 
138  miles.  The  wires  are  No.  00  B  &  S  and  are  spaced  96  inches 
apart.  Compute  the  capacitance  of  two  of  these  wires  if  used 
as  line  and  return. 

Prob.  28-13.  What  is  the  capacitance  of  a  40-mile  two- 
wire  transmission  line,  if  the  total  inductance  is  100  milli- 
henries? 

Prob.  29-13.  What  is  the  capacitance  between  two  wires 
of  the  154-mile  Big  Bend-Oakland  transmission  line?  The 

wires,  No.  000  B  &  S,  are 
spaced  10  feet  apart. 


fl 


128.     Application    of     a 
Sinusoidal    Voltage    to    a 
Condenser.     In    Fig.    231, 
A    is   an    alternator  which 
FIG.  231.     A  represents  an  alter-     supplies  a  variable  voltage 
nator  supplying  a  variable  volt-     to  ft  condenser  C.     We  will 
age  to  the  condenser  C.  ,  .   .  , , 

neglect    resistance    m    the 

circuit,  assuming  it  to  be  very  small.     The  voltage  of  the 
alternator  varies  sinusoidally  in  accordance  with  the  equation 

e  =  E  sin  ut,  (102) 

and  is  shown  in  Fig.  232. 

The  values  of  t  at  which  the  voltage  becomes  zero  are 
marked  on  the  time  axis. 

Since  there  is  no  resistance  in  the  circuit,  the  back  e.m.f. 
of  the  condenser  must  at  every  instant  equal  the  applied 


DIELECTRICS 


473 


voltage  after  conditions  in  the  circuit  have  reached  a  steady 
condition;  that  is, 

E  sin  <»t  =  g  A  (ft.  (103) 


FIG.  232.     The  voltage  of  the  alternator  A  in  Fig.  231 
varies  according  to  this  sine  curve. 


FIG.  233.  The  current  flowing  in  the  circuit  of  Fig.  231  is  represented 
by  this  cosine  curve.  Note  that  the  current  has  reached  its  maxi- 
mum value  at  the  instant  when  the  voltage  was  zero.  The  current 
thus  leads  the  voltage.  Note  that  the  time  scale  in  this  figure  is  dif- 
ferent from  that  of  Fig.  232. 


Differentiate  and  we  have 

i  —  EC  co  cos  co£ 


(104) 


as  an  expression  for  the  current  flowing.     This  curve  is 
plotted  in  Fig.  233. 


474      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Several  interesting  things  can  be  noted  from  this  curve. 
First,  the  current  also  varies  harmonically,  but  it  comes  to 
its  maximum  value  sooner  than  the  voltage.  In  fact,  the 
current  is  a  maximum  when  the  voltage  is  zero.  We  state 
this  by  saying  that  the  condenser  draws  a  leading  current. 
Also  the  magnitude  of  the  current  is  proportional  to  the 
applied  voltage,  to  the  capacitance  and  to  the  frequency,  for 
co  is  equal  to  2  TT  times  the  frequency  of  the  alternator  in 
cycles  per  second. 


FIG.  234.  The  curve  of  power  in  the  circuit  of  Fig.  231.  This  is 
also  sinusoidal,  but  note  that  the  frequency  is  double  that  of  the 
voltage  and  the  current. 

The  power  in  the  condenser  circuit  is  given  by 

p  =  ei  =  E2C  co  sin  orf  cos  orf  (105) 

CE2 

=  — 2~  ^  sin  2  co£.  (106) 

This  is  plotted  in  Fig.  234. 

The  power  therefore  is  also  a  sinusoidal  function  of  the 
time,  but  it  varies  at  double  frequency. 

The  energy  in  the  condenser  at  any  instant  is  given  by 


W=Cpdt.  (107) 


DIELECTRICS  475 

At  time  t  =  ^—  ,  we  have  for  the  voltage 
Z  co 

e  =  Esmu£-=E;  (108) 

2  co 

that  is,  at  this  instant  the  condenser  is  charged  to  maximum 
potential.  The  energy  in  the  condenser  at  this  instant  is 
given  by 

W=  fapdt  (109) 

Jo 

CE* 
—  jr—  co  sin  2  (t)t 


4  co 
which  gives 

W  =  ^f-  (110) 

This  is  the  expression  we  found  by  a  different  method  for 
the  energy  stored  in  a  charged  condenser. 
Note,  however,  that 


(Zpdt 

Jo 


(111) 


as  can  be  seen  by  inspection  of  Fig.  234,  for  the  positive 
and  negative  loops  are  equal  in  area.  That  is,  the  en- 
ergy stored  in  the  condenser  in  the  first  half  cycle  is  all 
returned  to  the  generator  during  the  next  half  cycle. 
This  assumes,  of  course,  no  resistance  and  no  losses  in  the 
condenser.  Under  these  conditions,  there  will  be  no  heating 
of  the  condenser,  for  there  is  no  energy  loss  in  it. 

Prob.  30-13.  A  sinusoidal  e.m.f.  of  150  volts,  maximum,  at 
a  frequency  of  60  cycles  per  second  is  applied  to  a  condenser  of 
5  mf.  capacitance 

(a)  What  maximum  current  flows? 

(6)  Plot  voltage,  current  and  power  curves  on  same  axes. 


476     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  31-13.  A  condenser  is  made  up  of  1200  sheets  of 
mica  0.010  centimeter  thick  and  leadfoil  plates  having  an  active 
area  of  20  X  18  centimeters.  If  an  alternating  voltage  with 
sinusoidal  wave  form  of  150  volts  (maximum)  at  a  frequency 
of  60  cycles  is  applied  to  the  terminals,  what  maximum  cur- 
rent will  flow? 

Prob.  32-13.  What  will  be  the  values  of  the  current,  volt- 
age and  power  in  the  circuit  of  Prob.  31-13  when  t  =  0.005 
second? 

129.  Condenser  Losses,  Dielectric  Hysteresis.  In  the 
above  treatment,  we  have  assumed  a  perfect  condenser,  — 
that  is,  one  which  does  not  leak,  has  no  resistance  in  its  leads 
and  no  losses  in  its  dielectric.  Condensers  constructed  of 
glass  or  mica  with  metal  plates  can  be  made  nearly  perfect. 
No  losses  are  apparent  in  such  a  condenser  unless  the  fre- 
quency is  high.  Paper  condensers,  or  condensers  formed 
of  sections  of  cables,  are  much  less  perfect.  There  are 
many  places  where  condensers  are  used,  or  where  there  is  a 
'condenser  action,  in  which  there  are  very  considerable  losses. 
These  are  important  not  only  on  account  of  the  cost  of  the 
lost  energy,  but  also  because  of  local  heating.  We  have  seen 
that  temperature  greatly  affects  the  properties  of  insulators, 
a  rise  in  temperature  always  lessening  the  resistance  and 
lowering  the  flash-over  point. 

A  cable  acts  as  a  condenser.  In  an  ocean  cable,  the  con- 
ducting wire  is  one  plate  of  the  condenser,  the  insulating 
material  is  the  dielectric,  and  the  sheath  of  the  cable  or  the 
sea  water  forms  the  other  plate.  Also  cables  are  often  used 
for  power  transmission  where  the  voltage  used  is  30,000 
volts  or  less,  particularly  in  cities.  The  transmission  wires 
are  assembled  together  with  impregnated  paper  insulation 
between  them,  the  whole  being  covered  with  a  lead  sheath  to 
keep  out  moisture.  In  such  an  arrangement  any  local  heat- 
ing is  very  serious,  as  it  is  likely  to  lead  to  break  down  of  the 
cable. 

There  are  three  sorts  of  losses  in  condensers:  first, 
that  due  to  the  resistance  of  the  leads,  which  is  usually 


DIELECTRICS  477 

small;  second,  the  loss  due  to  leakage,  which  in  the  case 
of  a  good  insulator  is  also  small;  third,  the  loss  in  the 
dielectric  itself.  This  last  is  most  important. 

We  meet  here  a  phenomenon  known  as  dielectric  hysteresis 
and  also  an  effect  due  to  absorption  or  "  soaking  into  "  the 
dielectric.  When  a  steady  voltage  is  applied  to  a  paper 
condenser,  careful  measurement  will  show  that  the  charge  will 
soak  into  the  condenser  for  a  considerable  length  of  time. 
Similarly  it  will  gradually  soak  out  on  discharge.  If  a 
charged  paper  condenser  is  discharged,  and  then  left  for  a 
short  interval,  a  further  small  voltage  will  appear  on  it,  and 


FIG.  235.     The  power  curve  of  a  condenser  having  losses.     Note  that 
the  upper  and  lower  loops  are  not  equal  as  they  are  in  Fig.  234. 

it  may  be  discharged  again.  Dielectric  losses  may  be  ex- 
plained in  either  of  two  ways.  The  first  is  to  assume  that 
the  electrons  of  the  dielectric  experience  a  certain  viscosity 
in  their  motion.  This  is  true  dielectric  hysteresis.  The 
second  is  that  the  dielectric  consists  of  a  large  number  of 
layers  with  high  resistances  between  layers.  Probably 
both  of  these  effects  are  present  in  many  cases.  The  former 
exerts  greatest  influence  at  high  frequencies.  The  latter  is 
the  cause  of  the  principle  loss  in  cable  insulation. 

If  a  condenser  with  losses  is  connected  to  an  alternator 
and  the  curve  of  power  plotted,  it  will  appear  as  in  Fig.  235. 
The  positive  and  negative  loops  are  no  longer  equal.  Of 


478     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

the  energy  supplied  to  the  condenser,  not  all  is  returned 
during  the  next  quarter  cycle.  Some  of  it  is  lost  in  heat. 
The  expression 

2ir 

p  dt  (112) 


C 
I 


is  no  longer  equal  to  zero,  but  is  twice  the  difference  in  area 
between  a  positive  and  a  negative  loop  of  the  power  curve. 
The  value  of  this  integral  is  equal  to  the  energy  loss  in  joules 
per  cycle  in  the  condenser. 

Prob.  33-13.  A  sinusoidal  voltage  of  100  volts  maximum  is 
impressed  across  the  terminals  of  a  condenser  having  a  capacity 
of  1.7  m.f.  At  60  cycles  frequency  due  to  losses  the  current 
flowing  is  represented  by  the  equation 

i  =  Em&x  Cw  cos  (orf  —  5°). 

Plot  the  current,  voltage  and  power  curves  on  the  same  axes. 

Prob.  34-13.  Either  by  using  a  planimeter,  by  counting 
squares  or  by  integrating,  determine  the  average  power  used  up 
in  overcoming  the  losses  in  the  condenser  of  Prob.  33-13. 

130.  Condensers  in  Parallel  and  Series.     If  two  conden- 

sers, of  capacitances  Ci  and 
C2,  are  connected  in  par- 
allel as  shown  in  Fig.  236, 
they  act  simply  as  a  larger 
condenser  of  capacitance 

FIG.  236.     Two    condensers   con-  (jp  =  Ci  +  C2.    (113) 

nected  in  parallel. 

This  is  easily  shown  in  the 

following  manner.  If  e  is  the  voltage  across  the  pair,  the 
currents  in  them  are  respectively 


The  total  current  is 


DIELECTRICS  479 

but  this  is  the  same  current  as  would  be  taken  by  a  single 
condenser  of  capacitance 

CP  =  Ci  +  C,. 

When  the  condensers  are  connected  in  series  as  in  Fig. 
237,  the  condition  is  different. 


FIG.  237.     Two  condensers  connected  in  series. 

If  a  charge  Q  is  passed  into  the  lead  to  the  condensers, 
the  same  charge  Q  appears  on  each.  The  voltages  to  which 
they  are  charged  will  be  respectively 

£         and         «. 

Ci  C2 

The  voltage  across  the  combination  will  then  be 


This  same  voltage  would  be  produced  if  the  pair  of  condensers 
were  replaced  by  a  single  condenser  of  capacitance  Cs,  so 
chosen  that 


that  is, 


€2 


This  expression  hence  gives  the  equivalent  capacity  of  the 
two  condensers  in  series. 

The  laws  for  combining  condensers  are  thus  the  same  as 
for  combining  conductances.  To  obtain  the  capacitance 
of  condensers  in  parallel,  add  the  individual  capacitances. 
To  obtain  the  capacitance  of  condensers  in  series,  take  the 


480      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

reciprocal  of  the  sum  of  the  reciprocals  of  individual  capaci- 
tances. 

When  condensers  are  connected  in  series,  the  smaller  con- 
denser is  subjected  to  the  higher  stress.  If  a  voltage  E 
is  applied  to  the  two  condensers  of  Fig.  237,  a  charge  will 
be  produced  on  each  of  value 

Q  =  CSE  =  „  V*^  coulombs.  (116) 

The  voltage  across  Ci  will  then  be 

Q 


and  across  €2 

Dividing  these  two  expressions, 


The  total  voltage  is  therefore  distributed  in  inverse  ratio 
to  the  capacitances. 

Prob.  36-13.  Three  condensers  have  respectively  capaci- 
tances of  2,  4  and  5  microfarads. 

(a)  What  is  the  capacitance  of  a  parallel  arrangement  of 
these  condensers? 

(6)  Of  the  2-microfarad  condenser  put  in  series  with  a  parallel 
arrangement  of  the  4-microfarad  and  5-microfarad  condensers? 

Prob.  36-13.     If  a  pressure  of  110  volts  is  impressed  across 
the  terminals  of  the  second  arrangement  in  Prob.  35-13, 
(a)  What  is  the  voltage  across  each  condenser? 
(6)  What  is  the  charge  in  each  condenser? 
(c)  What  is  the  total  charge  on  the  combination? 

Prob.  37-13.  It  is  desired  to  build  up  a  capacitance  of  2 
microfarads.  There  are  available  three  condensers  of  4,  1  and 
3  microfarads  respectively. 

(a)  Show  by  a  diagram  the  arrangement  of  these  condensers 


DIELECTRICS 


481 


which  will  produce  a  capacitance  most  closely  approximating 
2  microfarads. 

(6)  Compute  the  capacitance  of  your  arrangement. 

131.   Distribution  of  Stress  in  Insulation.     The  preceding 
section  has  considerable  bearing  on  the  design  of  insulation. 

Suppose,  for  example,  that  we  have  two  plates  as  in  Fig. 
238  separated  one  inch  and  subjected  to  an  applied  potential 
of  60,000  volts.  This 
voltage  is  about  all 
such  a  gap  will  stand 
when  the  plates  are 
separated  by  air,  even 
with  the  corners  well 
rounded  to  avoid  local 
high- voltage  gradients. 
Suppose  we  insert  a 
plate  of  glass  \  inch 
thick  of  dielectric  con- 
stant 5  as  shown,  in 
order  to  protect  against 
possible  breakdown. 
Examine  the  new  condi- 


FIG.  238.     Two  plates  separated  by  a 
layer  of  air  and  glass. 


tions. 

We  now  have  in  effect  two  condensers  in  series.  The  one 
formed  by  the  glass  has  a  capacitance  five  times  as  great  as 
the  one  formed  by  the  remaining  half  inch  of  air,  because 
of  the  high  dielectric  constant  of  the  glass.  Accordingly  if 
Eg  and  Ea  are  the  voltages  across  glass  and  air  respectively, 

Ea      1 


and  since 
we  obtain 


Eg  +  Ea  =  60,000, 

Eg    =     10,000, 

Ea  =  50,000. 


482     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


The  introduction  of  a  sheet  of  insulating  material  of  high 
dielectric  constant  thus  causes  most  of  the  stress  to  be  brought 
to  bear  on  the  remaining  material  of  low  dielectric  constant. 
The  above  discussion  assumes  no  leakage.  However,  a 
potential  of  50,000  volts  brought  to  bear  upon  one-half  inch 
of  air  will  surely  break  it  down.  A  spark  cannot  pass  clear 
across  on  account  of  the  glass,  but  the  air  will  be  ionized 
and  a  corona  will  form.  When  the  air  thus  becomes  con- 
ducting, nearly  all  of  the  60,000  volts  stress  will  be  brought 
to  bear  on  the  glass.  A  piece  of  glass  one-half  inch  thick  will 
surely  be  able  to  stand  this  voltage.  There  is  much  doubt, 
however,  if  it  can  stand  the  heating  due  to  the  continued 
formation  of  corona  if  the  potential  applied  is  alternating  so 
that  the  air  is  continually  being  broken  down.  Such  an  ar- 
rangement as  above,  while  it  would  be  safe  for  a  continually 
applied  potential,  would  therefore  be  likely  to  lead  to  diffi- 
culty if  the  supply  were 
alternating  and  of  the 
same  maximum  value. 

Prob.  38-13.  Two  plate 
electrodes  similar  to  those 
in  Fig.  238  are  placed  2 
inches  apart  and  a  poten- 
tial of  50,000  volts  applied. 
If  a  sheet  dielectric  \ 
inch  thick  and  having  a 
dielectric  constant  of  3  is 
inserted  between  the  elec- 
trodes, what  will  be  the 
new  distribution  of  poten- 
tial? 

Prob.  39-13.  The  three 
dielectric  sheets  shown  in 
Fig.  239  have  the  follow- 
ing dielectric  constants : 
A,  2.5;  B,  3;  C,  4.5.  What  is  the  potential  distribution  across 
the  different  sheets  when  60,000  volts  are  maintained  between 
the  plates  x  and  y? 


FIG.  239.  The  plates  x  and  y  are 
separated  by  three  dielectric  sheets 
A,  B  and  C. 


DIELECTRICS 


483 


Prob.  40-13.  Assuming  the  same  data  as  in  Prob.  39-13, 
compute  the  voltage  distribution  in  Fig.  239 : 

(a)  With  A  removed  and  B  and  C  as  in  the  figure, 
(6)  With  B  removed  and  A  and  C  as  in  the  figure, 
(c)  With  C  removed  and  A  and  B  as  in  the  figure. 

Prob.  41-13.  What  dielectric  sheet  or  sheets  of  arrangement 
in  Prob.  40-13  must  be  removed  to  produce  a  condition  most 
likely  to  cause  a  break  down  of  the  air?  Assume  the  dielectric 
sheets  to  have  much  greater  dielectric  strength  than  air. 

132.  Cylindrical  Con- 
denser. A  conductor, 
surrounded  by  insula- 
tion and  a  sheath,  as 
shown  in  Fig.  240,  forms 
a  cylindrical  condenser. 
This  is  the  arrangement 
of  a  single  -  conductor 
cable  used  in  power  trans- 
mission or  submarine  tel- 
egraphy. The  capacity 
of  such  a  condenser  may 
be  found  as  follows. 

Consider  an  elemen- 
tary cylinder  of  length 
one  centimeter  perpen- 
dicular to  the  paper, 
thickness  dx,  at  a  radius 

x.    The  capacity  of  such  a  cylinder  can  be  found  from  the 
formula  for  a  flat-plate  condenser, 


FIG.  240.  A  conductor  surrounded  by 
an  insulation  and  a  sheath  forms  a 
condenser. 


where 


KA   ,  .       , 
a—j — abfarads, 
dx 


K  is  the  dielectric  constant  of  the  material, 

1 


(120) 


a  is  the  constant 


36  TT  1020 


484      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

The  dimensions  are  in  centimeters;  A  is  the  cross-sectional 
area  of  the  dielectric.  Since  we  are  considering  unit  length 
of  the  cable, 

A  =  2  TTX. 

Now  the  total  condenser  can  be  considered  to  be  made  up 
of  these  elementary  condensers  in  series.  In  the  case  of 
condensers  in  series  we  have  seen  that  to  obtain  the  reciprocal 
of  the  total  capacitance,  we  must  find  the  sum  of  the  re- 
ciprocals of  the  individual  capacitances.  Inserting  the 
value  of  A,  we  have 

1     =       dx 
Cox       2irxaK' 

To  obtain  the  total  capacity  we  must  integrate  this  re- 
ciprocal to  cover  all  the  elementary  condensers. 

Hence 

B-rras"      «2» 

that  is, 

-•  -4 

rz 


or 

C  =  *^.  (123) 


Inserting  the  value  of  a,  we  have,  if  dimensions  are  in  centi- 
meters, the  formula  for  the  capacitance  of  the  cable  in  ab- 
farads  per  centimeter  length 

1  ~K 

C  =  10  x  i  Q2Q  --  abfarads  per  centimeter,      (124) 

'  log-S 

or,  since 

1  abfarad  =  109  farads 

=  1015  microfarads, 

1  7^ 

C  =  ^g  x  i  Q5  -  microfarads  per  centimeter,     (125) 


DIELECTRICS  485 


or  changing  to  common  logarithms, 


7x- 

0.0388  -      -  microfarads  per  mile.         (126) 


The  distribution  of  stress  in  the  cable  can  be  found  as 
follows.  The  voltage  across  each  elementary  condenser  is 
inversely  proportional  to  its  area  since  each  receives  the 
same  charge.  This  gives 

£=*,  (127) 

dx      x 

where  b  is  a  constant  to  be  determined.     Integrating, 

E=  r26^=610gep;  (128) 

,/r!        *  ri 

or 

6  =  -^-  (129) 


where  E  is  the  voltage  applied  to  the  cable. 

Hence  for  the  potential  gradient  at  any  point,  we  have 

i-  =  —     — volts  per  centimeter.  (130) 

d/*C  i         7*2 

e7*l 

This  distribution  of  stress  is  plotted  in  Fig.  241.     We  ob- 
tain the  maximum  stress  when  x  =  r\. 


~r  )      =  —      -  volts  per  centimeter.          (131) 

CLX/rnsaf  i  7*2 


It  is  this  value  of  stress  which  must  be  considered  in  cable 
design. 

Prob.  42-13.  What  is  the  capacitance  to  sheath  per  mile  of 
a  rubber-covered  single-conductor  cable  having  a  wire  of  No.  0000 
(B  &  S)  and  insulation  -fa  inch  thick? 


486      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  43-13.  If  impregnated  paper  is  substituted  for  the 
rubber  in  the  cable  of  Prob.  42-13,  what  will  be  the  capacitance 
per  mile?  The  dielectric  constant  of  impregnated  paper  may 
be  taken  as  2.5. 


FIG.  241. 


The  distribution  of  stress  in  the  insulation  of  a  cable  with 
a  metal  sheath. 


Prob.  44-13.  Describe  with  diagrams  how  an  ungrounded 
break  in  an  insulated  cable  could  be  located  provided  the 
capacitance  of  the  whole  cable  before  the  break  was  known. 

Prob.  45-13.  If  a  1000-volt  potential  is  applied  to  the  cable 
of  Prob.  42-13,  plot  the  potential  gradient  throughout  the 
insulation. 


SUMMARY    OF  CHAPTER   XIII 

AN  INSULATING  MATERIAL  is  one  which  conducts  no 
appreciable  current  when  only  moderate  voltages  are  applied 
to  it.  Glass,  porcelain,  rubber,  oil  etc.,  are  insulators  under 
ordinary  conditions. 

THE  DIELECTRIC  STRENGTH  of  an  insulating  material 
is  the  potential  gradient  necessary  to  break  it  down  and  allow 
a  current  to  flow.  Temperature,  freedom  from  moisture 
and  duration  of  voltage  are  some  of  the  factors  affecting  di- 
electric strength  of  a  material. 

CONDENSER  ACTION  is  the  name  given  to  the  phenomenon 
of  the  flow  of  current  on  to  or  off  from  a  plate  or  conductor 
when  its  potential  is  raised  or  lowered.  It  is  explained  by 
assuming  that  while  the  electrons  cannot  leave  the  plate,  they 
are  moved  a  little  in  position  and  are  under  stress.  The  ratio 
of  the  current  flowing  to  the  rate  of  change  of  voltage  is  the 
CAPACITANCE  of  the  condenser. 

A  CONDENSER  HAS  ONE  MICROFARAD  of  capacitance 
when  a  voltage  change  of  one  volt  per  second  causes  a  flow  of  a 
microampere,  or  when  a  change  of  one  volt  causes  a  change  of 
one  microcoulomb  in  the  charge  on  the  condenser. 

THE  DIELECTRIC  CONSTANT  of  a  material  is  the  number 
which  shows  the  ratio  of  the  capacitance  of  a  condenser  using 
that  material  as  a  dielectric  to  a  similar  condenser  using  air. 

THE  CAPACITY  OF  A  PARALLEL- PL  ATE  CONDENSER 
can  be  found  from  the  equation 

°i<r  A 
C  =  0.8842  X  10~7     -  microfarads. 

THE  CHARGE  ON  A  CONDENSER  is  found  by  the  equa- 
tion 

Q  =  CE. 

CAPACITANCE  MAY  BE  MEASURED  by  the  bridge  method 
or  directly  by  a  calibrated  ballistic  galvanometer, 

487 


488     PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

THE  CURRENT  ENTERING  OR  LEAVING  a  condenser 
through  a  resistance  when  a  voltage  is  suddenly  applied  or 
removed  is  represented  by  the  following  equation  : 


THE  TIME  CONSTANT  of  a  circuit  containing  resistance 
and  capacity  is  the  time  in  which  the  current  would  reduce 
to  zero  if  it  continued  to  decrease  at  the  initial  rate,  and  equals 
RC. 

THE  CHARGE  AT  ANY  TIME  ON  A  CONDENSER  which 
is  being  charged  is 

__t_ 
q  =  EC  (1  -  e    RC  ) 

and  while  being  discharged  is 

_t_ 
q  =  ECe-RC. 

THE  ENERGY  STORED  IN  A  CONDENSER  is  represented 
by  the  equation 

w 


THE  ENERGY  USED  UP  IN  THE  SERIES  RESISTANCE 
in  charging  a  condenser  from  a  constant  voltage  source  re- 
gardless of  the  amount  of  resistance  also  equals 


THE  MECHANICAL  FORCE  acting  between  the  plates  of 
a  parallel-plate  condenser  is  expressed  by  the  equation 

CE2 

f  =  -r^-  X  107  dynes 
2o 

or  by 

1        D2A 


f 


9  X  102°87rK 


in  which  D  is  the  electrostatic  flux  density  and  K  the  dielectric 
constant. 


DIELECTRICS  489 

THE  ELECTROSTATIC  FORCE  F  MAY  BE  MEASURED 
in  volts  per  centimeter  and  is  analogous  to  the  magnet- 
ization force  H,  measured  in  gilberts  per  centimeter.  It  is 
also  called  the  POTENTIAL  GRADIENT.  The  potential 
between  two  conductors  of  a  condenser  is  the  integral  of  the 
potential  gradient  between  the  conductors. 

THE  CAPACITANCE  OF  TWO  LONG  PARALLEL  AERIAL 
CONDUCTORS  is  expressed  by  the  equation 

0.0194 
C  =  -    — =—  microfarads  per  mile  of  line. 

logf 

WHEN  A  SINUSOIDAL  VOLTAGE  IS  APPLIED  to  a 
perfect  condenser,  a  sinusoidal  current  flows  which  reaches  its 
maximum  value  a  quarter  cycle  before  the  voltage  does.  If 

e  =  Emax  sin  coT 
then  i  =  CEmaxw  cos  wt. 

NO  POWER  IS  ABSORBED  BY  A  PERFECT  CONDENSER 
but  if  there  is  LEAKAGE  or  HYSTERESIS  loss  in  the  di- 
electric, then  power  is  consumed  and  the  temperature  rises. 

THE  CAPACITANCE  OF  CONDENSERS  IN  PARALLEL 
equals 

Cp  =  Ci  +  Cz  +  Cs 

THE  CAPACITANCE  OF  CONDENSERS  IN  SERIES 
equals 

1 


Cs 


JL       JL       JL 
Ci      C2      Cs 


THE  DISTRIBUTION  OF  STRESS  IN  INSULATION  is 
INVERSELY  PROPORTIONAL  to  the  dielectric  constant 
of  the  various  series  layers  of  material  composing  the  insulator. 

THE  CAPACITY  OF  AN  INSULATED  CABLE  may  be 
found  from  the  equation 

l£ 
C  =  0.0388 -  microfarads  per  mile. 


490      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

THE   MAXIMUM    STRESS   IN   A   CABLE   occurs    when 
x  =  ri  and  can  be  found  from  the  equation 


(—  -  ) 
<*' 


volts  per  centimeter. 


THE  FOLLOWING  RELATIONS   exist  among  the  units 
of  the  various  systems  used  in  this  chapter. 
1  statvolt  =  300  volts 

1  volt  =  108  abvolts 
1  abcoulomb  =  10  coulombs 

1  coulomb  =  3  X  109  statcoulombs 
1  statcoulomb  =  2.1  X  109  electrons 
1  abfarad  =  109  farads 

1  farad  =  106  microfarads 
1  microfarad  =  9  X  105  statfarads 


PROBLEMS   ON   CHAPTER   XIII 

Prob.  46-13.  A  circular  disk  of  glass  J  inch  thick  and  6 
inches  in  diameter  has  an  electric  potential  applied  between 
circular  plates  centrally  located  on  opposite  faces.  Assume  the 
surface  resistivity  to  be  2  X  1010  ohm-centimeters.  Assume 
also  that  the  glass  has  a  dielectric  strength  of  105  volts  per 
centimeter  but  that  10,000  volts  per  centimeter  surface  gradient 
will  flash  it  over.  Neglect  fringing  of  the  lines  of  electrostatic 
force  in  the  material. 

(a)  If  the  circular  electrodes  are  1  inch  in  diameter,  at  what 
potential  will  the  disk  break  down  and  how? 

(6)  If  the  electrodes  are  3  inches  in  diameter,  at  what  po- 
tential will  the  disk  break  down  and  how? 

Prob.  47-13.  A  sheet  of  tinfoil  2  X  24  inches  and  0.001 
inch  thick  is  placed  in  the  center  of  a  sheet  of  paper  3  X  26 
inches  and  0.0015  inch  thick,  and  having  a  dielectric  constant 
of  4.  t  The  tinfoil  is  then  covered  with  another  sheet  of  paper  of 
the  same  size  as  the  bottom  one,  and  another  similar  sheet  of 
tinfoil  is  laid  on  the  top  sheet  of  paper.  The  whole  is  carefully 
wrapped  about  a  mandrel  of  J  inch  diameter,  forming  a  cylin- 
drical condenser  3  inches  long  and  slightly  over  j  inch  in  diam- 
eter. Connections  are  now  led  to  the  two  sheets  of  tinfoil. 
What  is  the  capacitance  of  the  condenser? 

Prob.  48-13.   What      *A  AA 

will  be  the  capacitance     ' rT      VVV  I 


B, 


fl 


found  as  in  Prob.  47- 

13  if  we  initially  pile 

up  12  sheets  of  tinfoil 

2  X  240    inches    and 

0.001  inch  thick,  and 

12  sheets  of  3  X  242    FIG.  242.     A  circuit    containing    resistance 

inch  paper  of  0.0015  and  capacitance. 

inch  thickness? 

Prob.  49-13.  The  capacitance  of  the  condenser  in  Fig. 
242  is  4  microfarads,  the  constant  voltage  of  the  battery  E  is 
20  volts,  ^i  is  30  ohms,  R2  is  45  ohms. 

(a)  Write  the  equation  of  the  transient  current  delivered 
by  the  battery  E  upon  closing  the  switch  S. 

491 


492      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

(6)  What  current  will  be  flowing  0.0005  second  after  closing 
the  switch, 

(1)  in  fl2? 

(2)  in  #1? 

(3)  in  C? 

Prob.  50-13.  (a)  What  will  be  the  charge  in  the  condenser 
of  Fig.  242,  0.08  second  after  the  switch  is  closed? 

(6)  What  will  be  the  voltage  across  C,  R2  and  Ri  at  this 
instant? 

Prob.  61-13.  After  the  condenser  of  Prob.  49-13  has  become 
fully  charged,  the  switch  S  is  opened.  What  current  will  be 
flowing  through  Ri,  0.0002  second  after  the  switch  is  opened? 

Prob.  62-13.  What  charge  will  there  be  on  the  condenser  of 
Prob.  51-13,  0.0002  second  after  the  switch  is  opened? 

Prob.    63-13.       In 

S     |  J_     Fig.    243,    Ci    has    a 

capacitance  of  2.4  mf. 
and  is  charged  to  a 
potential  of  20  volts. 
#1  =  10  ohms  and 
FIG.  243.  The  switch  S  connects  the  posi-  #2=6  ohms.  C2  has 
tively  charged  terminals  of  the  condensers.  a  capacitance  of  4.6 

mf.  and  is  charged  to 

a  potential  of  8  volts.  When  the  switch  is  thrown  it  connects 
the  positive  sides  of  the  condensers  together.  What  current 
will  be  flowing  0.07  second  after  the  switch  is  closed? 

Prob.  64-13.  If  the  initial  polarity  of  one  of  the  condensers 
in  Prob.  53-13  is  reversed,  what  current  will  be  flowing  0.07 
second  after  the  switch  is  closed? 

Prob.  65-13.  It  will  be  noted  that  the  charge  and  discharge 
current  curves  of  condensers  through  resistance  are  identical. 
The  same  is  true  even  if  the  resistance  is  inductive.  On  the 
basis  of  this  fact,  prove  that  in  charging  condensers  from  a 
constant  source  of  potential  through  an  inductive  resistance, 
just  half  the  energy  input  is  lost  in  heat. 

Prob.  56-13.  If  the  dielectric  of  a  condenser  has  a  constant 
high  resistivity,  and  if  the  applied  potential  is  maintained  con- 
stant, will  the  maximum  potential  gradient  in  the  dielectric  be- 
come greater  or  less  with  time  after  application  of  constant 
potential? 


DIELECTRICS 


493 


Prob.  57-13.  Explain  the  phenomenon  of  dielectric  ab- 
sorption on  the  basis  of  leakage  in  the  dielectric. 

Prob.  58-13.  Four  condensers  each  of  1  microfarad  capaci- 
tance are  connected  in  series.  One  of  the  end  condensers  has  a 
conductance  of  10~8  mhos,  the  others  have  zero  conductance. 
Plot  the  potential  across  each  condenser  as  a  function  of  the 
time  after  applying  a  potential  of  1000  volts  to  the  series. 

Prob.  59-13.  Repeat  Prob.  58-13  using  conductances  of 
10~8,  2  X  10-8,  4  X  10~*  and  8  X  10~8  mhos  respectively  for  the 
condensers  in  series. 


\ 


*      .  -     .  M  j 

FIG.  244.  The  current  and  voltage  curves  of  a  leaky  condenser. 
The  curve  with  the  small  waves  in  it  is  the  current  curve.  Prof. 
F.  S.  Dellenbaugh. 

Prob.  60-13.  Construct  the  power  curve  from  the  curves  of 
current  and  voltage  shown  in  Fig.  244.  The  frequency  is  60 
cycles  per  second;  the  voltage  scale,  250  volts  per  inch;  the  cur- 
rent scale,  1.8  ampere  per  inch.  What  is  the  average  power  loss 
in  the  condenser  under  these  conditions? 


APPENDIX 


TABLE  I 

RESISTIVITY  AND  TEMPERATURE  COEFFICIENT 


Material 

Resistivity  in 
microhm-centi- 
meters at  20°  C. 

Temperature  co- 
efficient per  degree 
C.  per  ohm  at 
20°  C. 

Aluminum 

2  828 

0  003°) 

Antimony 

36  to  46 

0  0039 

Bismuth 

119 

0  0045 

Brass 

8  7 

0  0010 

Constantin   . 

50 

0  000005 

Copper 

Annealed  Intnl.  Standard  

1.724 

0.00393 

Hard  drawn  

1  77 

0  00382 

Pure  

1  68 

0  0041 

German  Silver  

30 

0  00036 

Gold  

2  2  to  2  26 

0  0037 

Iron,  Commercial  

11  to  13  5 

0  0055 

Hard  cast  

98 

Lead  . 

20  1  to  21  4 

0  0042 

Manganin  . 

42  to  74 

0  00003 

Monel  Metal 

42  6 

0  00198 

Mercury  

96 

0  0009 

Nickel  

12  to  14 

0  006 

Platinum  

9  7  to  16  5 

0  0038 

Platinium-iridium  .  . 

24  6 

0  0012 

Silver  

1  64  to  1  85 

0  0040 

Steel,  Hard.... 

47  2 

0  0016 

Soft  

17  4 

0  0042 

Rail  

13  8  to  21  6 

Tantalum  .  .  . 

15  3 

0  0027 

Tin  

10  4  to  12  5 

0  0043 

Tungsten  

7  6 

0  0039 

Zinc  

6  04  to  fi  ^fi 

0  0040 

Advance  Metal  .  . 

48  8 

0  000018 

/a/a  

49  0 

0  000005 

Superior  

86  4 

0  00081 

Nichrome  .... 

QQ  fi 

0  00044 

Nichrome  II  

109  5 

0  00016 

Calorite  

119  5 

494 


APPENDIX 


495 


02 


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-  §  § 


PQ    d  .!3 

-II 


I1 


02    tf 

*!l 


Mi 


II* 


I 


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21* 

«  s 


^'^^o   ddddd   ddddd   ddddd 


00  »— i  CO  CO  CO     00  TJH  OS  OO  <N 


i-H  O5  C35  C5      00  i-H  CO  00 


i  CO     C<l  <N  »-i 


°°t>'l>'(:oir;> 


«5  C>  00  *O          CO  O5  CO -^  C<|     -^  O  O5  CO  5J 


OO  W  O5  00  S 


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(N  CO     i— I  CO  GO  IO 
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0000  OOOOO     OOOOO     i-HT-i 


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CO  CO  t>-  i—  i     CNOO>OOCO     CN  *O  >-H 


O5  C<|  CO  i—  i  10 

00  ^^  C^l  00  C^ 
iCOCOCNO 


8-^  O  CO          O  CO  N  i— i  Tfi     CQ  00  O5  CO  O5     -"T  O  CO 
cooqoq        coco-^coos    pc<|^'^oo    t>-oooi 


|O 


O          i-H 


486      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 

Prob.  43-13.  If  impregnated  paper  is  substituted  for  the 
rubber  in  the  cable  of  Prob.  42-13,  what  will  be  the  capacitance 
per  mile?  The  dielectric  constant  of  impregnated  paper  may 
be  taken  as  2.5. 


FIG.  241. 


The  distribution  of  stress  in  the  insulation  of  a  cable  with 
a  metal  sheath. 


Prob.  44-13.  Describe  with  diagrams  how  an  ungrounded 
break  in  an  insulated  cable  could  be  located  provided  the 
capacitance  of  the  whole  cable  before  the  break  was  known. 

Prob.  46-13.  If  a  1000-volt  potential  is  applied  to  the  cable 
of  Prob.  42-13,  plot  the  potential  gradient  throughout  the 
insulation. 


SUMMARY    OF  CHAPTER   XIII 

AN  INSULATING  MATERIAL  is  one  which  conducts  no 
appreciable  current  when  only  moderate  voltages  are  applied 
to  it.  Glass,  porcelain,  rubber,  oil  etc.,  are  insulators  under 
ordinary  conditions. 

THE  DIELECTRIC  STRENGTH  of  an  insulating  material 
is  the  potential  gradient  necessary  to  break  it  down  and  allow 
a  current  to  flow.  Temperature,  freedom  from  moisture 
and  duration  of  voltage  are  some  of  the  factors  affecting  di- 
electric strength  of  a  material. 

CONDENSER  ACTION  is  the  name  given  to  the  phenomenon 
of  the  flow  of  current  on  to  or  off  from  a  plate  or  conductor 
when  its  potential  is  raised  or  lowered.  It  is  explained  by 
assuming  that  while  the  electrons  cannot  leave  the  plate,  they 
are  moved  a  little  in  position  and  are  under  stress.  The  ratio 
of  the  current  flowing  to  the  rate  of  change  of  voltage  is  the 
CAPACITANCE  of  the  condenser. 

A  CONDENSER  HAS  ONE  MICROFARAD  of  capacitance 
when  a  voltage  change  of  one  volt  per  second  causes  a  flow  of  a 
microampere,  or  when  a  change  of  one  volt  causes  a  change  of 
one  microcoulomb  in  the  charge  on  the  condenser. 

THE  DIELECTRIC  CONSTANT  of  a  material  is  the  number 
which  shows  the  ratio  of  the  capacitance  of  a  condenser  using 
that  material  as  a  dielectric  to  a  similar  condenser  using  air. 

THE  CAPACITY  OF  A  PARALLEL- PL  ATE  CONDENSER 
can  be  found  from  the  equation 

C  =  0.8842  X  10~7-jT  microfarads. 

THE  CHARGE  ON  A  CONDENSER  is  found  by  the  equa- 
tion 

Q  =  CE. 

CAPACITANCE  MAY  BE  MEASURED  by  the  bridge  method 
or  directly  by  a  calibrated  ballistic  galvanometer, 

487 


498      PRINCIPLES  OF  ELECTRICAL  ENGINEERING 


TABLE  V 
SPECIFIC  HEAT  OF  VARIOUS  MATERIALS 


Material 

Specific 
heat  in 
calories  per 
gram 

Material 

Specific 
heat  in 
calories   per 
gram 

Aluminum 

0  0217 

Sulphur  

0.173 

Antimony 

0  0489 

Tin  

0.0551 

Bismuth 

0  0302 

Tungsten  

0.0336 

Bromine 

0  107 

Zinc  

0.0935 

Cadmium 

0  0560 

Glass 

Calcium 

0.1520 

crown  

0.161 

Cobalt 

0.103 

flint  

0.117 

Copper.  . 

0.0931 

Ice  

0.530 

Gold  

0.0316 

Rubber  

0.481 

Iron 

Mica  

0.206 

cast 

0  119 

Paraffin 

0  694 

wrought 

0  115 

Porcelain 

0  26 

Lead 

0  031 

Brass 

0  088 

Manganese 

0  107 

German  Silver 

0  095 

Mercury  

0.0335 

Alcohol 

Nickel      . 

0  109 

ethyl 

0  648 

Platinum 

0  0323 

methyl 

0  601 

Silver.  ... 

0  0559 

Petroleum 

0  511 

TABLE  VI 

DIELECTRIC  CONSTANTS  (Average  Values) 


Material 

Dielectric 
constant 

Material 

Dielectric 

constant 

Ebonite  . 

2  7 

A.cetone 

26  6 

Glass 
flint  
hard  crown  .  , 

9.9 

7  0 

Alcohol  (0°  C.) 
amyl  
ethyl 

17.4 
28  4 

lead  

6.6 

methyl  

35 

Guttapercha  

4  1 

Ammonia 

22 

Mica  

5  8 

Benzene 

2  3 

Paraffin  

2  I 

Glycerine 

56  2 

Shellac.. 

3  1 

Petrole  um 

2  1 

Air  

1 

^^ater  (pure) 

81 

INDEX 

The  figures  refer  to  page  numbers. 


Abampere,  22,  204 

Abcoulomb,  467 

Abfarad,  437 

Abhenry,  235 

Abohm,  23 

Absolute  system,  22,  147 

Absorption  of  condenser  charge, 

477,  489 
Abvolt,  22,  250 
Abvolt-second,  141 
Accuracy  of  computations,  231 

of  measurements,  55,  97 
Acid,  definition  of,  110 
Aging  of  magnets,  318,  320 
Air,  flux  density  in,  188, 190,  204 

number  of  free  electrons  in,  408 

permeability  of,  148 
Air-cored  solenoid,  207,  230 
Air  gap,  magnetic  pull  in,  278 
Air  gaps,  163,  239 

effects  of,  164,  166,  225,  237 
Alexanderson  alternator,  309 
Alkaline  battery,  120,  123 
Alloys,  resistivity  of,  78,  100 
Alternating  current,  definition  of, 

20 

Alternating-current  meters,  prin- 
ciples of,  375 

Alternating-current     transformer, 
170 

computation  of  magnetic  circuit 
of,  170 

corrections  in  computation  of, 
174 


Alternator,  elementary,  333 

high-speed,  343 

large,  345 

low-speed,  343 

rotating  field  of,  344 

stationary-armature,  345 
Alternator   supplying   condenser, 
472 

power  curve,  474 

voltage  and  current  curves,  473 
Aluminum,  conductivity  of,  80 

cost  of,  130 

manufacture  of,  130 

resistivity  of,  78,  101 

transmission  lines  of,  89,  101 
Aluminum  Mfg.  Plant,  129 
Ammeter,  38 
Ampere,  definition  of,  18,  43 

international,  19 
Ampere-hour,  120 
Ampere's  law,  136 
Ampere-turns  and  flux,  155,  158 
Amplification,  distortionless,  403 
Amplifier,    thermionic,   400,    403, 
426 

grid  characteristics  of,  402 
Analogy  of  electric  and  magnetic 

fields,  461,  470 
Anode,  109,  126 
Arcs,  421,  427 

characteristic  of,  422,  424 

conduction  through,  152,  192 

effect  of  magnetic  field  on,  425 

incandescent  spot  of,  421,  425 
Arcing  of  switches,  424 

danger  point  of,  425 


499 


500 


INDEX 


Arc-light  circuits,  422 
Arc  welding,  423 
Armature,  connections,  336 

current  paths,  337,  380 

windings,  339 

Armature  core,  flux  in,  226,  229 
Armature  current,  of  motor,  382, 

384,  386 

Atomic  weight,  114 
Atoms,    in    one    gram-molecule, 
116,  133 

in  solutions,  107,  109 
Auxiliary  reactions,  116 
Averages,  use  of,  220,  222 

B 

Back    electromotive    force,    369, 

380,  382,  383 
Balanced  load,  68 
Balancer  set,  68 
Ballistic  galvanometer,  140,  215, 

307,  375,  386,  446 
calibration  of,  215,  244 
Base,  definition  of,  110 
Battery  combinations,  46 
Bauxite,  130 
B-H  curves,  154,  303 
Bonds  of  electrons,  435 
"  Booster  "  feeder,  47 
Breakdown  of  air,  418,  420,  421, 

482. 
Breakdown     of    insulators,    432, 

465,  482 

Brush,  commutator,  336 
Brush  discharge,  420 


Cable,  cross-section  of,    98 

distribution  of  stress  in,   485, 

490 
Cables,  condenser  action  of,  476 


Calcium  carbide,  manufacture  of, 

130 

Calcium  chloride,  110 
Calorie,  61,  70 
Capacitance,  436,  487 

measurement  of,  444,  446,  447, 

448,  449,  487 

of  insulated  cable,  485,  489 
of  long  aerial  conductors,  468, 

470,  472,  489 
of  parallel-plate  condenser,  439, 

487 

Carbon  arcs,  422 
Carborundum,  130 
Carrying  capacity,  92 
Cast  iron,  magnetic  properties  of, 

153,  154 

Cathode,  109,  126 
dark  space,  411 

Caustic  soda,  manufacture  of,  130 
C.  g.  s.  system,  22, 147 
Charges,    positive   and   negative, 

107 

Chemical  change,  107 
effects,  108,  109 
groups,  110 
potential,  112 
voltage,  126 

Chlorine,  manufacture  of,  130 
Circuit-breaker,  magnetic,  293 
Circuit  with  capacitance  and  re- 
sistance, 454,  491 
Circular  field,  189 
Circular  mil,  77 
Clark  cell,  21 

Close  coupling  of  coils,  281,  289 
Coefficient  of  coupling,  281 
of  hysteresis  loss,  315,  318 
of  mutual  induction,  280,  284, 

289 
Coefficient  of  self-induction,  255, 

256,  280,  284,  287 
Coercive  force,  300,  305,  314 


INDEX 


501 


r 


Coil  in  magnetic  field,  370,  371, 

378 
Coils  in  slots,  225,  226,  227,  228 

of  round  and  rectangular  wire, 

224 

Commutator,  335,  336,  362,  379 
Commutatorless  generator,  346 
Complex  ions,  110 
Computation,    exact   vs   approxi- 
mate, 220,  222,  225,  231 
Concentrated  coil  winding,  343 
Concentration  of  electrolyte,  117, 

121,  124,  126 
Concrete    structures,    electrolysis 

of,  128,  133 
Condenser,  252,  434,  436 

action,  434,  476,  487 

air  and  glass,  481 

charge,  440,  443,  488 

charge  curve,  453 

charging    current    curve,    451, 
454 

charging  through  a  resistance, 
448,  458 

circuit,  power  in,  474 

curve  of  leaky,  493 

dielectric  absorption  of,  477 

discharge,  453,  455,  488 

discharge  current,  455 

discharge  curve,  456 

energy  relations,  456,  488 

hydraulic  analogy  of,  441 

losses,  476 

measurement  of  current,  437 

mechanical  force  on,  459,  488 

parallel-plate,  439,  459,  487 

potential  gradient,  461 

simple,  435 

sinusoidal  voltage  on,  472 

three-dielectric,  482 

time  constant  of,  451,  488 
Condensers,  in  parallel,  478 

in  parallel  and  series,  478 


Condensers,  in  parallel  and  series, 
capacitance  of,  478,  481,  489 
in  parallel  and  series,  total  vol- 
tage of,.  480 
in  series,  479 
two  connected,  492 
Conductance,  definition  of,  35,  43 
Conduction,  electrolytic,  107,  109, 

392 

in  X-ray  tubes,  408,  410 
metallic,   resume"   of  character- 
istics, 391 

non-metallic,  392,  426 
thermionic,  393,  394,  395,  396, 

397,  426 
Conductivity,  of  electrolytes,  111, 

132 

of  materials,  79 
vs  permeability,  146 
Conductor,    chemical   change  in, 

107 

force  on,  367,  386 
in  iron  sheath,  191,  233 
in  iron  trough,  232 
in  moving  field,  340 
local  currents  hi,  348 
solid  cylindrical,  200 
tubular,  199 
Contactor,  294 
Contactors,      alternating-current, 

354 
Continuous  current,  definition  of, 

19 

Coolidge  X-ray  tube,  406 
fluorescence  of,  407 
focussing  shield  of,  407 
potential  of,  407 

Copper,  conductivity  of,  80,  100 
refining,  126 

refining,  cost  of,  126,  127 
refining,  energy  required,  127 
resistance  of,  77 
resistivity  of,  78,  80,  100 


502 


INDEX 


Copper-clad  steel  wire,  91,  101 
Copper-wire  tables,  85 
Core  loss,  309 

total,  359,  363 

Core  of  portable  instruments,  373 
Corona,  420,  427,  482 

loss,  392,  420 
Correction  for  "  lay,"  89 
Coulomb,  definition  of,  18 

number  of  electrons  in,  115 
Critical  potential  for  spark  dis- 
charge, 416,  420 
Critical     pressure    of    discharge 

tubes,  414 

Crystal  rectifier,  393 
Cumulative  discharge,  412,  426 
Current,  capacity,  safe,  92,  101 

caused  by  chemical  change,  107 

curve  of  rotating  coil,  379 

growth  and  decay  of,  263,  271, 
288 

in  dielectrics,  435 

in  X-ray  tubes,  408,  410 

leakage  of,  98 

measurement  of,  37,  38,  43 

non-sinusoidal,  316 

per  volt,  35 

persistence  of,  271 

relations  in  networks,  29 

resume  of  mechanism  of,  431 

sheet,  208,  223 

through    condenser    in    series 
.  with  resistance,  448,  458,  487 

types  of,  19 

unit  of,  18 

Cutting  lines  of  force,  331, 362, 363 
Cylindrical  condenser,  482 

D 

De  Forest,  thermionic  device,  400 
Dielectric,  absorption,  477,  489 
constant,  438,  487 


Dielectric,  hysteresis,  476,  477 

strength,  variation  of,  433 
Dielectrics,  units  used  for,  490 
Direction,  positive  vs  negative,  108 
Disconnecting  switch,  388 
Displacement  current,  436,  438 
Dissociation,  109,  110,  111 
Distribution  of  stress  in  insula- 
tion, 481,  489 


E 


Earth,  field  of,  234 

resistance  of,  127 

stray  currents  in,  128 
Eddy-current  loss,  347,  349,  355, 
363 

analysis  of,  350 

in  short-circuited  rings,  354 

variation  of,  357,  359 
Eddy  currents,  252,  302,  309,  318, 
347,  349,  363 

direction  of,  348 

in  a  ring,  350 

in  iron  sheet,  355 

reduction  of,  349 
Edgewood,    Md.,   chlorine  plant, 

131 

Edison  storage  battery,  120,  123, 
133 

capacity  of,  125 

charge  and  discharge  curves,  125 

condition  of,  124 

electrolyte  for,  124 

intermediate  reaction  of,  125 

reaction  of,  124 

voltage  drop  in,  125 
Efficiency  of  a  motor,  382 
Electric,  arcs,  152 

condensers,  434 

circuit,  17,  21,  136,  138 

current,  resume  of  mechanism, 
431 


INDEX 


503 


Electric,  energy,  definition  of,  60 

force,  54 

furnace  products,  130 

inertia,  261,  269 

potential    gradient,    461,    468, 
470,  489 

power,  definition  of,  58 

power,  measurement  of,  58,  70 
Electrical, engineering,  basis  of,  136 

engineers,      classification      and 
qualifications,  9,  13 

transmission,   summary  of  ad- 
vantages, 5 

Electricity,    relation    to   magnet- 
ism, 136 
Electrochemical  equivalent,    115, 

133 

Electrochemical  processes,  129, 133 
Electrode,  109 
Electrolysis,  127 

damage  by,  128,  133 

of  fused  salts,  129 

with  alternating  currents,  128 
Electrolyte,  effect  of  dilute,  123 

heating  of,  126 

specific  gravity  of,  121,  124 
Electrolytes,  107,  109 
Electrolytic,cells,  voltage  of,  112 

conduction,  107,  109 

copper,  126 

copper,  cost  of,  127 

copper,  impurities  of,  126 

iron,  magnetic  properties  of,  319 

refining,  126,  133 
Electromagnet,  U-shaped,  158 
Electromagnetic  units,  274 
Electromotive  force,  138,  328 

and  JR  drop,  27 

chemical,  263 

positive  and  negative,  29 

short-circuited,  288 

thermal,  263 

unit  of,  20 


Electron  bonds,  435 
Electron  '  theory,  17 
Electrons,  107,  109,  132,  269,  391, 
396,  404,  408,  431 

concentration  of,  408 

cumulative  discharge  of,  412 

emission  of,  408,  421 

evaporation  of,  396,  397 

flow  of,  144 

in  condensers,  441 

in  dielectrics,  431 

light  effect  of,  407,  410,  411 

mean  free  path  of,  409 

ratio  of  charge  to  mass,  405 

velocity  of,  405,  407,  411,  426 

viscosity  of  motion  of,  477 
Electroplating,  108,  126 

amount  of  deposit,  114 
Electrostatic  field,  416,  418,  465, 
466,  467 

flux  density,  466 

forces,  459,  470,  489 

lines  of  force,  417,  465 

voltmeter,  464,  465 
Elementary  alternators,  333 
Energy,  in   condenser,  457,  458, 
459,  474,  475 

in  air  gap  of  cast-steel  ring,  276 

input,  272 

of  magnetization,  304 

stored  in  magnetic  field,   270, 

289 

Engineering  accuracy,  231 
Ergs,    stored    in   magnetic   field, 

274,  288 
Euler's  lines  of  flux,  142 


Factory- wiring  plan,  46 
Farad,  437 

Faraday,  definition  of,  115,  133 
number  of  electrons  in,  115, 133 


504 


INDEX 


Faraday's  laws,  114, 117, 127,  132, 
136 

lines  of  flux,  143 
Feeder,  47 
Field,  parallel,  uniform,  234 

strength  of  long  wire,  201 

windings,  343,  385 
Filaments  of  flux,  238,  240 
Fixation  of  nitrogen,  130 
Flaming  arcs,  422 
Flash-over,  419 

on  insulator,  419 
Fleming's.right-hand  rule,  331,  367 

thermionic  valve,  393 
Fluorescence,  407 
Flux,  251,  253,  350,  354 

amount  of,  146 

and  ampere- turns,  155 

and  current,  direction  of,  165, 
189 

current  required,  156 
Flux  density,  143,  148,  152,  176, 
188,  232,  234,  235,  237,  242, 
304 

alternating,  305 

at  center  of  circular  form,  202, 
205 

in  iron  sheath  of  cable,  190 

lagging,  301,  321 

maximum,  315 

measurement  of,  307,  375 

near  wire  in  air,  190,  196 

of  circular  coil,  202,  203,  204, 
205 

of  rectangular  coil,  194 

of  solenoid,  211,  214,  215 

relation   to   eddy-current    loss, 
354 

relation  to  magnetizing  force, 
153 

relation  to  permeability,  152 

variation  in,  313,  359 

wave  form,  253,  254 


Flux,  diagrams,  237 

distribution,  187,  197,  202,  225 

distribution   in   leakage   paths, 
173 

in  air  gaps,  165 

in  armature  core,  226,  229 

lines,  142, 145 

lines,    direction    of,    165,    189, 
235,  236 

lines   and   mutual    inductance, 
283 

linkage,  255 

linkage,    change   in,    328,    347, 
362,  363 

measurement  of,  138,  142,  216 

of  induction  coil,  256 

paths,  224,  231,  240 

total,     between     two     parallel 

wires,  243 
Foot-pound,  53,  60 
Force,on  a  condenser,  459 

on  a  conductor,  367,  380 

on  a  conductor,  amount  of,  368 

on  meter  coils,  373 

on  wire  in  a  field,  368,  370,  378 
Four-pole  generator,  175 

armature  of,  339 

frame  of,  340 
Frequency,  very  high,  309 
Fringing,  173,  238,  239,  277 
Froelich's  equation,  310,  321 


Galvanometer,  ballistic,  140,  215, 
244,  307,  375,  386,  446 

sensitive,  375 
Gas  on  electrodes,  113 

amount  liberated,  114 
Gas  pressure  in  X-ray  tubes,  409 
Gaseous  discharge  tube,  410 

X-ray  tube,  409 
Gases,  electrons  in,  408 


INDEX 


505 


Gases,   identification   of  by   dis- 
charge tubes,  411 
Gasket  joints,  128 
Gassing  of  Edison  batteries,  125 
Gauss,  144,  148,  176 

equivalent  of,  155 
Generator,  328,  330 

direct-current,  335 

high-current,  low- voltage,  346 

vs  motor,  367,  380 

voltage  diagram,  79 

without  commutator,  347 
Gilbert,  147,  176 
Gilberts  per  centimeter,  149 

equivalents  of,  155 
Glass,  dielectric  strength,  432 

resistivity  of,  95 
Gram-molecule,  115 

number  of  atoms  in,  116,  132 
Grid  characteristics  of  amplifiers, 

402 

Grounds,  resistance  of,  128 
Growth  and  decay  of  current,  260, 
261,  262,  263,  265,  266,  296 


Hard  vacuum,  407,  412 
Harmonical  variation, of  flux,  253, 

287 

of  voltage  and  current,  473 
Heat,  energy,  61 
losses,  61,  62,  66,  70 
treatment,  318,  322 
Henry,  definition  of,  255,  257 
Homopolar  generator,   346,   347, 

362,  380 
Horse  power,  53 
House-wiring  diagram,  93 
Hydraulic  analogies,   17,   18,   19, 
20,  21,  52,  53,  142,  145,  441, 
442 


Hydrochloric  acid,  resistivity  of, 

111 

Hydrogen,  manufacture  of,  130 
Hydroxyl  ions,  110 
Hysteresis, coefficient,  315,  318 

dielectric,  476 

effect  of  frequency  on,  308,  359 

magnetic,  298,  301 
Hysteresis  loop,  302, 304, 306, 307, 
310,  313,  317,  321 

plotting  of,  307 

unsymmetrical,  316 
Hysteresis  loss,  301,  306,  321 

in  dynamo  armatures,  302 

in  transformers,  302,  308 

measure  of,  306 

mechanical  analogy  of,  301 

of  electrical  machinery,  348 

Steinmetz  equation  for,  321 

variation  of,  314,  359 


Idaho  Power  Co.'s  Plant,  3 
Ignition  system,  252 
Impregnated  paper,  resistivity  of, 

95 

Impressed  voltage,  effect  of  sud- 
den change  in,  267,  288 
Induced  voltage,  250,   257,   281, 

328,  362 

of  moving  coils,  330,  362 
Inductance,   255,   256,   261,   264, 

269,  283 

Induction  coil,  251 
action,  252 
uses  of,  252 

voltage  obtainable,  252 
Induction  motor,  379 
Inductive    circuits,    current    and 
time  relation,  263,  271,  288, 
377 


506 


INDEX 


Inductive  circuits,  instantaneous 
conditions  in,  259,  262,  288 

transients  in,  258 

with  resistance,  258,  297 
Inertia,  electric,  261,  264,  269 
Instantaneous  values,  60 
Insulating  materials,  431,  487 

resistivity  of,  95 
Insulation    of    laminations,    349, 

357 
Insulation  resistance,  97,  101 

test  diagram,  50 
Insulator  breakdown,  418 

with  petticoat,  434 
Insulators,  design  of,  433 

effect  of  air  pockets  in,  433 
Interior  wiring,  sizes  for,  92, 101 
Interlinking    circuits,    145,    176, 

261 

Internal  resistance,  117,  120,  122 
Interrupter,  thermionic,  399 
lonization,  107,  407,  408,  410 

cumulative,  413 

velocity,  414 
Ions,  107,  132,  408,  413 

complex,  110 

direction  of  motion  of,  108,  109 
IR  drop,  27,  176 
PR  loss,  62,  66 


Joule,  60,  70 


Kilowatt,  52,  53 
Kilowatt-hour,  60,  70 
Kinetic  energy,  269,  272,  288 
Kirchhoff's  laws,  25,  43,  258 

applied    to    magnetic    circuits, 
168,  183,  240 

diagram,  25,  27,  29 


Kirchhoff's    laws,   hydraulic   an- 
alogy of,  diagram,  26 


Lamination    of:  cores,   349,   357, 

363 

Lay  of  strands  in  cable,  89 
Lead  battery,  120,  133 

charge  and  discharge  curves,  122 

chemical  reaction  of,  121 

condition  of,  121 

disadvantages  of,  123 

internal  resistance  of,  122 

voltage  of,  121 
Lead-sheathed  cable,  98,  101 
Lead  storage  cell,  121 

charge  and  discharge  curve,  122 
Leakage  current,  98,  167 

flux,  167,  168,  171,  177,  223 

from  transmission  lines,  420 

lines,  169 

of  insulating  materials,  96,  101 

path,  cross-section  of,  173 
Left-hand  rule,  367,  386 
Lenz's  law,  251,  287 
Lifting  magnets,  137,  278,  279 
Light  effect  of  electrons,  408,  410 
Lightning  arresters,  392,  418 
Line  integral,  184,  189,  200,  226, 
242 

graphic  example.  185 

law,  183 

of  flux  paths,  187 

zero  value,  186,  188,  199,  223, 

236 

Line  losses,  66,  70 
Line,  unit  of  flux,  176 
Lines,  of  electrostatic  stress,  417, 

465 

of  flux,  143 
Linkage,  146,  261 

change  in,  250,  255,  281 


INDEX 


507 


Lithium  hydrate,  124 
Load  center,  94 

effect  on  motor  speed,  384 
Local  action,  118 
Long-distance  telephony,  400 
Loss  of  energy  in  magnetic  field, 
272 


M 

Magnetic,  blow-outs,  425 

chuck,  181,  294,  295 
Magnetic  circuit,  136,  137 

computation  of,  159,  160,  163, 
166,  170,  174,  183,  225,  229, 
240 

energy  consumed  in,  138,  176 
flux  distribution,  187 
heating  of,  300,  306 
of  dynamo,  168 
of  three-phase  transformer,  160, 

161 

non-uniform,  151,  159 
simplified  computation,  166 
with  air  gaps,    163,    177,   225, 

233,  237,  277,  304 
Magnetic    circuits,     two    forms, 

224 

Magnetic  energy,  270,  272,  289 
Magnetic  field,  136,  183,  373 
applicability  of  formulas,  274 
energy  stored  in,  272,  274,  275, 

289,  302,  321 
inside  a  solid  conductor,  200, 

243 
inside  a  tubular  conductor,  199, 

200 

mechanical  analogy  of,  272 
near  a  straight  wire,  193.  243 
of  armature  core,  226,  245 
of  circular  conductor,  202,  205, 

243 
of  conductor  in  a  slot,  224 


Magnetic  field,  of  long  transmis- 
sion lines,  188,  195,  201,  232, 
243 

of  solenoid,  207,  244 
of  toroid,  218,  245 
uniform,    introduction   of   iron 

into,  230 

Magnetic  flux,  137  / 

effects  of,  139 
measurement  of,  138 
total,  145 
Magnetic  flux  lines,  142 

tendency  of,  276 
Magnetic,insulation,  167,  177 
leakage,  168,  177 
lines,  iron  from  air,  236,  237, 

239 

poles,  240 

potential  gradient,  461,  468 
properties    of    iron    and    steel, 
153,  154,  298,  315,  317,  318 
pull,  276,  278,  289 
pull  in  air  gaps,  277 
vs  electric  circuits,  138,  144,  149, 
152,  160,  164,  167,  172,  176, 
188,  232,  240 

Magnetism,      relation     to     elec- 
tricity, 136 
Magnetization  curves,    162,    177, 

299,  304,  312 
mean,  310 

Magnetization  of  permanent  mag- 
nets, 320 
Magnetizing  force,  149,  193,  204, 

233,  242 

and  current,  191 
relation  to  flux  density,  153 
Magneto,  319 
Magnetomotive ,  force,    138,    145, 

147,  148,  176,  187,  242 
forces,  value  of,  173 
Manganin,  85 
Maxwell,  141,  143,  147,  176 


508 


INDEX 


Mean  free  path  of  electrons,  409, 

410,  426 

Mean  magnetization  curves,  310 
Mean  self-inductance,  281 
Mercury-arc, lamp,  422 

rectifiers,  392,  423 
Metallic    conduction,    resume    of 

characteristics,  391 
Metals,  refining  of,  126 
Meter,  direct-current,  372 

cores  and  poles  of,  373 
Meter  scale,  evenly  divided,  373 
Meters,  371,  375,  386 
Mho,  35 

Mho-centimeter,  80 
Mica,  dielectric  strength  of,  432 
Microfarad,  437,  487 
Microhm-centimeter,  76 
Mil-foot  wire,  77 
Mobility  of  ions,  111,  116,  132 
Moisture  on  insulators,  419 
Molecules,  107,  110,  407 
Moore  light,  411 
Motors,  principles  of,  377 

slowing  down  of,  384 

speed  of,  381,  383,  384,  386 
Moving  field,  340 
Multipolar, field,  343 

generator,  terminal  voltage  of, 

339 
Mutual  induction,  280,  289 

N 

Negative  feeders,  128 
Nernst  "  glower,"  85 
Networks,  computation  of,  32 
Neutral  wire,  68 
Nickel-iron  battery,  120,  123 

reaction  of,  124 
Nitrogen,  fixation  of,  130 
Non-metallic  conduction,  392 
Normal  solution.  Ill 


Norway  iron,  318 
No-voltage  release,  230 


Oersted,   definition  of,    14d,   147, 

176 

Oersted's  discovery,  136 
Ohm,  definition  of,  21,  43 
Ohm-centimeter,  76,  100 
Ohm's  law,  22,  43,  146,  392 
Ohm's  law,  application  of,  23,  155 
to  inductive  circuits,  260 
to  insulating  materials,  96 
Ohm's  law,  and  electric  arcs,  152 
for  magnetic  circuits,  146,  155, 

176,  219,  240,  275 
Ohms  per  mil-foot,  77,  100 
Oil  switch,  230 
Oscillating  current,  definition  of, 

20 
Oscillator,  C.  G.  Smith's,  413 

thermionic,  400,  403 
Oscillogram,  454,  493 
Out  of  phase,  254 
Output  of  a  motor,  382 
Oxygen,  manufacture  of,  130 


Parallel  conductors,  flux  distribu- 
tion, 195,  197,  198 
Parallel-plate  condenser,  439,  487 
Parallel  wires,  charges  on,  470 
Pasted  type  of  cell,  120 
Period  of  alternating  flux,  352 
Permanent  magnets,  300,  319,  322 
Permeability,   analogy  to  dielec- 
tric constant,  438 

definition  of,  146,  177 

greater  than  unity,  148,  168 

infinite,  235 

of  saturated  iron,  298 


INDEX 


509 


Permeability,  of  vacuum,  168 

reciprocal  of,  235 

relation  to  flux  density,  152 

variation  of,  152,  298 

vs  conductivity,  146 
Permeameter,  307 
Petticoats,  96,  433 
Phosphorus,  manufacture  of,  130 
Plante  type  of  ceU,  120 
Plate    current,    thermionic,    402, 

403     . 
Pliotron,  characteristic  curves  of, 

402 

Polarization,  118 
Pole  face,  effect  of  area,  279 
Pole,  shoes,  168 

strength,  240,  279 
Poles,    demagnetizing    effect    of, 

300 

Polyvalent  atoms,  110 
Porcelain,  dielectric  strength   of, 

432 

Positive  vs  negative  direction,  108 
Potassium  hydroxide,  124,  133 

manufacture  of,  130 
Potential,  difference,  26 

energy,  272 

gradient,  111,  149,  404 

series,  111,  112,  132 

series,  care  in  the  use  of,  112 
Power,  chief  sources  of,  2 

computation  diagram,  54 

consumed  by  resistance,  55,  70 

curve  of  condenser  having  losses, 
477 

equation,  52,  55,  70 

equation,  application  of,  54 

in  condenser  circuit,  474 

input,  270,  272,  369 

loss  curve,  352 

losses,  57 

measurement  of,  58 

plants,  location  of,  6 


Power,  synonyms  for,  26,  53 

transformed  by  a  motor,  382 

unit  of,  21 

Primary, batteries,  109,  111,  117, 
133 

batteries,  disadvantages  of,  118 

coil,  171,  251 
Proportionality  factor,   140,   146, 

148,  211,  368,  436 
Pulsating  current,  definition  of,  19 


Quadrature,  254 
Quartz,  manufacture  of,  130 
resistivity  of,  95 

R 

Radicals,  110 
Radio  detector,  393 
Rail  return,  128 
Recalescence  point,  318,  319 
Rectifier,  crystal,  392 

mercury-arc,  392 

Smith's,  413 

thermionic,  399 
Regulation,  62,  70 

of  transformer,  171 

of  transmission  lines,  188 
Relays  in  power  stations,  230 
Reluctance,  146 

drop,  176 

of  leakage  path,  170 
Reluctances,   series  and  parallel, 

150,  177 

Repeating  coil,  218 
Repulsion  motor,  379 
Residual    magnetism,    300,    305, 

314,  321 

Resistance,    computation   of,   34, 
76,  150 

drop,  27,  17$ 


510 


INDEX 


Resistance,  effect  on  inductance, 
264 

measurement  of,  40,  43 

of  electric  arcs,  152 

of  stranded  wire,  89 

of  wires,  simple  rule  for,  86 

per  mil-foot,  77,  100 

unit  of,  21 

Resistance-temperature  curve,  83 
Resistivity,  76 

of  insulating  materials,  95 

of  metals,  78 

relation  to  eddy-current  loss,  354 
Resistor  combinations,  34,  35,  36, 

44,51 

Retentivity,  217,  298,  300 
Reversible  reaction,  109,  120,  133 
Reversing  switches,  217 
Revolving   field,    advantages    of, 
343 

alternator,  341,  343 

motor,  379 

Richardson's  law,  395,  426 
Right-hand, rule,  331,  367 

screw,  166 

screw  relation,  165,  189 
Ring-shaped  magnet,  300 
Ripple  in  voltage,  338 
Rotor  of  vertical  alternator,  343 
Rubber,  resistivity  of,  95 

S 

San  Franciscito  plant,  2,  4 
Saturated  iron,  298,  300 
Saturation  current,  395,  397,  426 

computation  of,  398 
Secondary,  chemical  action,  120, 
128 

coil,  171,  251 

Self-cooled  transformer,  61 
Self-induction,  255 

voltage  of,  287 


Sheathed    insulated    conductors, 
condenser  effect  of,  483 

stress  in  the  insulation,  486 
Sheet  steel,  magnetic  properties 

of,  153,  154 
Short  circuit  of  storage  batteries, 

122 
Short    circuits    on    transmission 

lines,  90,  101 

Short-circuited    rings,    eddy-cur- 
rent loss  in,  354     . 
Shunt  motor,  385 
Silicon,  manufacture  of,  130 

steel,  315,  318,  358 
Silver,  recovery  of,  126 
Sinusoidal,  current,  253 

flux,  351 

flux  variation,  351 

voltage  on  condenser,  472 
Skin  effect,  202 
Slip-rings,  341 

Slots,  armature,  226,  229,  335 
Slotted  armature,  335 
Smith  rectifier  and  oscillator,  413 
Sodium,  manufacture  of,  130 
Solenoid,  air-cored,  207,  230,  244 

flux  density  of  field,  188,  207 

iron-cored,  230 

plunger  type,  230 

reluctance  of  long,  214,  244 
Solid  cylindrical  conductor,  200 
Space-charge  effect,  418 
Space  factor,  357 
Spark  discharge,    414,   415,  421, 
427 

critical  potential,  416,  420 
Spark-gap  settings,  417,  427 
Speed  of  motor,  381,  383,  384,  385, 

386 

Star  connection,  35 
Star-delta  connection,  51 
Statcoulombs,  467 
Stator  of  vertical  alternator,  343 


INDEX 


511 


Statvolts,  462 

Steel,     alloys,     manufacture    of, 
130 

castings,  magnetic  properties  of, 
153,  154 

effect  of  composition,  317,  322 

resistivity  of,  78 
Steinmetz  equation,  314,  321 

application   of,    315,   316 
Step-up  transformer,  253 
Storage  batteries,  109,   111,   119, 
120,  133 

buckling  of  plates,  122 

charge  of,  120,  121,  125 

discharge  of,  120,  121,  125 

over-discharge  of,  121 

rating  of,  120 

short  circuit  of,  122 
Storage-battery  curves,  122,  125 
Storage  battery  with  booster,  8, 

48 

Stranded  wire,  88,  101 
Stray  currents,  128,  133,  348 
Submarines,  detection  of,  235 
"Sulphating,"  122 
Superpower  system,  7 
Surface  resistivity,  96 
Swan  Falls  plant,  2 
Synchronous  motor,  379 


Telephone,  currents,  403 

receiver,  137 

receivers,  hum  in,  188 

repeater,  392,  400,  401,  426 
Temperature  coefficient,  80,   100 

negative,  85,  96,  100 

of  alloys,  84,  100 

of  copper,  82,  100 
Temperature  rise,  83,  100 
Thermionic  amplifier,  400,  403 

grid  characteristics  of,  402 


Thermionic  conduction,  393,  426 

analogy  to  vaporization,  396 

curves  of,  394,  395 

maximum  current,  397 

Richardson's  law  of,  395,  397, 

426 

Thermionic    current,    curves    of, 
194,  195,  402 

measurement  of,  394 
Thermionic,  Deforest's  device,  400 
Thermionic    emission,   392,    397, 
421 

effect  of  ionization  in,  408,  410 
Thermionic,  interrupter,  399 

oscillator,  400,  403,  404 

rectifier,  399 
Thermionic  tube,  392 

with  grid,  400 

Thermionic   valves,  393,  397 
Three-phase  alternator,  armature 
of,  341 

voltage  curve  of,  342,  343 

generator,  343 

transformer,     computation    of, 

160,  170 

Three-wire  system,  66,  70 
Time  and  current  relations,  263, 

268,  288 

Time    constant,    264,    266,    268, 
288 

of  condenser,  451,  488 
Tooth-tip  flux,  227 
Toroid,  definition  of,  218 

flux  in,  221,  245 

magnetic  field  of,  218,  220 
Toroidal  repeating  coil,  291 
Torque,  370,  379,  381,  386 
Transformer  67,  251,  253 

action,  251 

action  diagram,  171 

losses  in,  302 

oil,  433 

plot  of  hysteresis  loop,  308 


512 


INDEX 


Transients,  258 

in  inductive  circuits,  258 
Transmission,    efficiency    of,    62, 
70 

losses,  66,  70 

regulation  of,  64 

system  diagrams,  63,  65,  67,  69, 
72 

three-wire,  66 

voltages,  66 

Triangle  connection,  35 
Trolley-car   voltage-boosting    de- 
vice, 47 

Tubular  conductor,  199 
Tungar,  410 
Tungsten. filaments,  85 

steel,  315,  318,  319,  322 
Turbo-generator,  137 
Two-pole  d-c.  generator,  336 

voltage  curve  of,  338 
Two-pole,  field,  343 

motor,  163 


U 


Underwriters'  table,  92,  101 
Unidirectional  rotation,  378,  379 
Uniform    field,    introduction    of 

iron  into,  230,  234 
Unit  current,  204,  244 
Units, in  absolute  system,  22 

used  for  dielectrics,  490 
Univalent  atoms,  109 


Vacuum,  hard,  407,  412 
permeability  of,  168 
tubes,  9 

Valence,  114 

Variometer,  284 

"Varley  loop,"  42 
test,  diagram,  42 


Varnished  cambric,  resistivity  of, 

95 

Vectorial  addition,  196 
Velocity,  of  electrons,  405,  426 
moving    conductor,    333,    335, 

359,  381 

Volt,  definition  of,  21,  43 
Voltage,     chemically      produced, 

111 

distribution    in    inductive    cir- 
cuits, 287 

drop,  allowable,  93 
gradient,  96 
induced,  250,  281 
regulator,  284 
regulator,  induction,  284 
ripples  in,  338 
sinusoidal,  472 

Voltage  generated  by  moving  con- 
ductors, 328,  330,  340,  362, 
367,  381 

direction  of  motion,  331 
Voltage,     measurement     of,     38, 

43 
by      electrostatic      attraction, 

464 

by  spark  gaps,  417 
diagram,  39 
Voltage,  of  coil  in  air  gap,  329, 

330,  333,  334 
of  condensers   in   parallel  and 

in  series,  480 
of  lamp  combinations,  diagram, 

45 
of  self-induction,  255,  258,  280, 

287 

of  wire  in  air  gap,  332 
Voltages,   generated    by    moving 

field,  341 
in  motors,  382 
used  in  plating,  126 
Voltmeter,  39 
Volt-second,  140 


INDEX 


513 


w 

Walker,    Dr.    Miles,  magnetizing 

curves,  298 
Water,  as  insulator,  107 

dissociation  of,  111 

formed  in  storage  batteries,  121 

pipes,  electrolysis  of,  128,  133 
Watt,  53,  70 
Watt-hour  meter,  358 

eddy  currents  in,  358 
Wattmeter,  59 
Wattmeter  connection  diagrams, 

59 

Watt-second,  60,  61,  70 
Weight  of  wire,  simple  rule  for, 

86 
Welding  by  arcs,  423 

of  steel  rings,  423 
Weston  voltmeter,  387 


Wheatstone  bridge,  40 

diagram,  40,  42,  49 

slide-wire,  42 
Wire  gauges,  85 
Working  flux,  335 
Wrought  iron,  magnetic  proper- 
ties of,  153,  154 


X-rays,  406,  426 
X-ray  tubes,  392,  404 

Coolidge's,  406 

effect  of  gas  pressure,  409,  410, 
411,  412 

old  type,  408,  409 


Y  connection,  35 


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